product rule algebra representation

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Product Rule

In Calculus, the product rule is used to differentiate a function. When a given function is the product of two or more functions, the product rule is used. If the problems are a combination of any two or more functions, then their derivatives can be found using Product Rule. The derivative of a function h(x) will be denoted by D {h(x)} or h'(x).

Product Rule Definition

The product rule is a general rule for the problems which come under the differentiation where one function is multiplied by another function. The derivative of the product of two differentiable functions is equal to the addition of the first function multiplied by the derivative of the second, and the second function multiplied by the derivative of the first function. The function may be exponential, logarithmic function , and so on.

Product Rule Formula

If we have a function y = uv, where u and v are the functions of x. Then, by the use of the product rule, we can easily find out the derivative of y with respect to x,  and can be written as:

(dy/dx) = u (dv/dx) + v (du/dx)

The above formula is called the product rule for derivatives or the product rule of differentiation .

In the first term, we have considered u as a constant and for the second term, v as a constant.

Product Rule Proof

Product rule can be proved with the help of limits and by adding, subtracting the one same segment of the function mentioned below:

Let f(x) and g(x) be two functions and h be small increments in the function we get f(x + h) and g(x + h).

Let F(x) = f(x)g(x) and F(x + h) = f(x + h)g(x + h)

Then, the derivative of a function is

By adding and subtracting f(x + h)g(x), we get

By using the definition of a derivative, we get

= f(x + 0) g’ (x) + g(x) f ‘(x)

F'(x) = f(x)g’ (x) + g(x)f ‘(x).

which is the derivative of two functions and is known as the product rule in derivatives .

Product Rule for Different Functions

The product rule for different functions such as derivatives, exponents, logarithmic functions are given below:

Product Rule for Derivatives:

d(uv)/dx = u(dv/dx)+ v(du/dx)

where u and v are two functions

Product Rule for Exponent:

If m and n are the natural numbers, then x n × x m = x n+m.

Product rule cannot be used to solve expression of exponent having a different base like 2 3 * 5 4 and expressions like (x n ) m . An expression like (x n ) m can be solved only with the help of Power Rule of Exponents where (x n ) m = x nm .

Product Rule for Logarithm:

For any positive real numbers A and B with the base a

where, a≠ 0, log a AB = log a A + log a B

Product Rule for Partial Derivatives:

If we have a function z = f(x,y) g(x,y) and we want to find out the partial derivative of z, then we use the following formula

Zero Product Rule:

Zero product rule states, the two non zero numbers are only zero if one of them is zero. If a and b are two numbers then ab = 0 only either a = 0 or b = 0.

if (x-1)x = 0, either x – 1 = 0 or x = 0

It means that if x – 1 = 0, then x = 1

Values of x are 0 and 1. They are also called roots of the equation. Mainly used to find the roots of equations, and it works if one side of the equation is zero.

Triple Product Rule:

Triple product rule is a generalization of product rule. If f(x), g(x) and h(x) be three differentiable functions, then the product rule of differentiation can be applied for these three functions as:

D[f(x). g(x). h(x)] = {g(x). h(x)} * D[f(x)] + {f(x). h(x)} * D[g(x)] + {f(x). g(x)} * D[h(x)]

Product Rule Example

Simplify the expression: y= x 2 × x 5

Given: y= x 2 × x 5

We know that the product rule for the exponent is

x n × x m = x n+m.

By using the product rule, it can be written as:

y = x 2 × x 5 = x 2+5

Hence, the simplified form of the expression, y= x 2 × x 5 is x 7 .

Differentiate y = sin x cos x

Given: y = sin x cos x

dy/dx = d(sinx cos x)/dx

While differentiating, it becomes

Differentiate the terms, dy/dx = sin x (-sin x) + cos x (cos x)

dy/dx = -sin 2.x + cos 2 x

dy/dx =cos 2 x – sin 2 x

By using identity,

dy/dx = cos 2x

Therefore, dy/dx = cos 2x

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Product Rule

Product rule in calculus is a method to find the derivative or differentiation of a function given in the form of the product of two differentiable functions. That means, we can apply the product rule, or the Leibniz rule, to find the derivative of a function of the form given as: f(x)·g(x), such that both f(x) and g(x) are differentiable. The product rule follows the concept of limits and derivatives in differentiation directly. Let us understand the product rule formula, its proof using solved examples in detail in the following sections.

What is the Product Rule?

Product rule in calculus is a method used to find the derivative of any function given in the form of a product obtained by the multiplication of any two differentiable functions. The product rule in words states that the derivative of a product of two differentiable functions is equal to the sum of the product of the second function with differentiation of the first function and the product of the first function with the differentiation of the second function. That means if we are given a function of the form: f(x)·g(x), we can find the derivative of this function using the product rule derivative as,

\(\frac{d}{dx}\) f(x)·g(x) = [g(x) × f'(x) + f(x) × g'(x)]

Product Rule Formula

We can calculate the derivative or evaluate the differentiation of the product of two functions using the product rule formula in Calculus. The product rule formula is given as,

\(\frac{d}{dx}\) f(x) = \(\frac{d}{dx}\) {u(x)·v(x)} = [v(x) × u'(x) + u(x) × v'(x)]

• f(x) = Product of differentiable functions u(x) and v(x)
• u(x), v(x) = Differentiable functions
• u'(x) = Derivative of function u(x)
• v'(x) = Derivative of the function v(x)

Derivation of Product Rule Formula

In the previous section, we learned about the product formula to find derivatives of the product of two differentiable functions. For any two functions, product rule may be given in Lagrange's notation as

(u v)' = u'·v + u·v' or in Leibniz's notation as

\(\dfrac {d}{dx}\) (u·v) = \(\dfrac {du}{dx}\)·v + u·\(\dfrac {dv}{dx}\)

Let us see the proof of the product rule formula here. There are different methods to prove the product rule formula, given as,

• Using the first principle
• Using chain rule

Product Rule Formula Proof Using First Principle

To prove product rule formula using the definition of derivative or limits, let the function h(x) = f(x)·g(x), such that f(x) and g(x) are differentiable at x.

⇒ h'(x) = \(\mathop {\lim }\limits_{Δx \to 0}\) [h(x + Δx) - h(x)]/Δx

= \(\mathop {\lim }\limits_{Δx \to 0}\) \(\frac{f(x+Δx)g(x+Δx) - f(x)g(x)}{Δx}\)

= \(\mathop {\lim }\limits_{Δx \to 0}\) \(\frac{f(x+Δx)g(x+Δx) - f(x)g(x+Δx) + f(x)g(x+Δx) - f(x)g(x)}{Δx}\)

= \( \mathop {\lim }\limits_{Δx \to 0} \frac{[f(x+Δx) - f(x)]g(x+Δx) + f(x)[g(x+Δx) - g(x)]}{Δx}\)

= \( \mathop {\lim }\limits_{Δx \to 0} \frac{[f(x+Δx) - f(x)]g(x+Δx)}{Δx} + \mathop {\lim }\limits_{Δx \to 0} \frac{f(x)[g(x+Δx) - g(x)]}{Δx}\)

= \( (\mathop {\lim }\limits_{Δx \to 0} \frac{[f(x+Δx) - f(x)]}{Δx})( \mathop {\lim }\limits_{Δx \to 0} g(x+Δx) )+ (\mathop {\lim }\limits_{Δx \to 0}f(x)) (\mathop {\lim }\limits_{Δx \to 0} \frac{[g(x+Δx) - g(x)]}{Δx})\)

= \( g(x)\mathop {\lim }\limits_{Δx \to 0} \frac{[f(x+Δx) - f(x)]}{Δx} + f(x)\mathop {\lim }\limits_{Δx \to 0}\frac{[g(x+Δx) - g(x)]}{Δx}\)

∵ \(\mathop {\lim }\limits_{Δx \to 0} \frac{[f(x+Δx) - f(x)]}{Δx}\) = f'(x) and \(\mathop {\lim }\limits_{Δx \to 0}\frac{[g(x+Δx) - g(x)]}{Δx}\) = g'(x)

⇒ \(\frac{d}{dx}\) f(x)·g(x) = [g(x) × f'(x) + f(x) × g'(x)]

Hence, proved.

Product Rule Formula Proof Using Chain Rule

We can derive the product rule formula in calculus using the chain rule formula by considering the product rule as a special case of the chain rule. Let f(x) be a differentiable function such that h(x) = f(x)·g(x).

\(\frac{d}{dx}\) (f·g) = [δ(fg)/δf][df/dx] + [δ(fg)/δg][dg/dx] = g(df/dx) + f(dg/dx)

Product Rule for Product of More Than Two Functions

The product rule can be generalized to products of more than two factors using the same product rule formula. For example, for three functions, u(x), v(x), and w(x), product given as u(x)v(x)w(x), we have,

\( \frac{d(uvw)}{dx}=\frac{du}{dx}vw+u\frac {dv}{dx}w+uv\frac {dw}{dx}{\frac {d(uvw)}{dx}}={\frac {du}{dx}}vw+u{\frac {dv}{dx}}w+uv{\frac {dw}{dx}}\)

How to Apply Product Rule in Differentiation?

In order to find the derivative of the function of the form h(x) = f(x)g(x), both f(x) and g(x) should be differentiable functions. We can apply the following given steps to find the derivation of a differentiable function h(x) = f(x)g(x) using the product rule.

• Step 1: Note down the values of f(x) and g(x).
• Step 2: Find the values of f'(x) and g'(x) and apply the product rule formula, given as: h'(x) = \(\frac{d}{dx}\) f(x)·g(x) = [g(x) × f'(x) + f(x) × g'(x)]

Let us have a look at the following example given below to understand the product rule better.

Example: Find f'(x) for the following function f(x) using the product rule: f(x) = x·log x.

Here, f(x) = x·log x

u(x) = x v(x) = log x

⇒u'(x) = 1 ⇒v'(x) = 1/x

⇒f'(x) = [v(x)u'(x) + u(x)v'(x)] ⇒f'(x) = [log x•1 + x•(1/x)] ⇒f'(x) = log x + 1

Answer: The derivative of x log x using the product rule is log x + 1.

☛ Topics Related to Product Rule:

• Applications of Derivatives
• Differential Equation

Examples on Product Rule

Example 1: Find the derivative of x· cos(x) using the product rule formula.

Let f(x) = cos x and g(x) = x.

⇒f'(x) = -sin x ⇒g'(x) = 1

⇒[f(x)g(x)]' = [g(x)f'(x) + f(x)g'(x)] ⇒[f(x)g(x)]' = [(x•(-sin x) + cos x•(1)] ⇒[f(x)g(x)]' = - x sin x + cos x

Answer: The derivative of x cos x using product rule is (- x sin x + cos x).

Example 2: Differentiate x 2 log x using the product rule formula.

Let f(x) = log x and g(x) = x 2 .

⇒f'(x) = (1/x) ⇒g'(x) = 2x

⇒[f(x)g(x)]' = [g(x)f'(x) + f(x)g'(x)] ⇒[f(x)g(x)]' = [(x 2 •(1/x) + log x•(2x)] ⇒[f(x)g(x)]' = x + 2x log x

Answer: The derivative of x 2 log x using product rule is (x + 2x log x).

Example 3: Apply the product rule to differentiate (1 - 2x)·sin x.

Let f(x) = (1-2x), g(x) = sin x

⇒g'(x) = cos x ⇒f'(x) = -2

⇒[f(x)g(x)]' = [g(x)f'(x) + f(x)g'(x)] ⇒[f(x)g(x)]' = [(sin x•(-2) + (1 - 2x)•(cos x)] ⇒[f(x)g(x)]' = - 2 sin x + cos x - 2x cos x

Answer: The derivative of (1 - 2x)·sin x using product rule is - 2 sin x + cos x - 2x cos x.

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Practice Questions on Product Rule

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FAQs on Product Rule

What is product rule of differentiation in calculus.

The product rule is one of the derivative rules that we use to find the derivative of functions of the form P(x) = f(x)·g(x). The derivative of a function P(x) is denoted by P'(x). If the derivative of the function P(x) exists, we say P(x) is differentiable, that means, differentiable functions are those functions whose derivatives exist. A function P(x) is differentiable at a point x = a if the following limit exists.

\(P'(x) = \mathop {\lim }\limits_{h \to 0} \frac{P(a+h)-P(a)}{h}\)

How to Find Derivative Using Product Rule?

The derivatives of the product of two differentiable functions can be calculated in calculus using the product rule. We need to apply the product rule formula for differentiation of function of the form, f(x) = u(x)v(x). The product rule formula is given as, f'(x) = [u(x)v(x)]' = [u'(x) × v(x) + u(x) × v'(x)] where, f'(x), u'(x) and v'(x) are derivatives of functions f(x), v(x) and u(x).

What is Product Rule Formula?

Product rule derivative formula is a rule in differential calculus that we use to find the derivative of product of two or more functions. Suppose two functions , u(x) and v(x) are differentiable , then the product rule can be applied to find (d/dx)[u(x)v(x)] as,

f'(x) = [u(x)v(x)]' = [u'(x) × v(x) + u(x) × v'(x)]

How to Derive the Formula for Product Rule?

Product rule formula can be derived using different methods. They are given as,

• Using derivative and limit properties or first principle

Click here to check the detailed proof of product formula.

How to Use Product Rule in Differentiation?

Product rule can be used in finding the differentiation of a function f(x) of the form u(x)v(x). Derivative of this function using product rule can be given as, f'(x) = [u(x)v(x)]' = [u'(x) × v(x) + u(x) × v'(x)]

How to Derive Product Rule Using Definition of Limits and Derivatives?

The proof of the product rule can be given using the definition and properties of limits and derivatives. For a function f(x) = u(x)·v(x), the derivative f'(x) can be given as,

What are Applications of Product Rule Derivative Formula? Give Examples.

We can apply the product rule to find the differentiation of the function of the form u(x)v(x). For example, for a function f(x) = x 2 sin x, we can find the derivative as, f'(x) = sin x·2x + x 2 ·cos x.

How to Find Derivative of Product of More Than Two Functions Using Product Rule?

We can generalize the formula for differentiation of products of more than two functions using the same product rule formula. For example, for three functions, u(x), v(x), and w(x), product given as u(x)v(x)w(x), we have,

How Can we Verify Quotient Rule Using the Product Rule in Calculus?

To prove the quotient rule using the product rule and chain rule, we can express the function f(x) = u(x)/v(x) as f(x) = u(x)•1/v(x) and further apply product rule formula to find f'(x) = (1/v(x))u'(x) - u(x)•(1/v(x)) 2 •v'(x) = \(\frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}\).

Product Rule

The product rule tells us the derivative of two functions f and g that are multiplied together:

(fg)’ = fg’ + gf’

(The little mark ’ means "derivative of".)

Example: What is the derivative of cos(x)sin(x) ?

We have two functions cos(x) and sin(x) multiplied together, so let's use the Product Rule:

(fg)’ = f g’ + f’ g

Which in our case becomes:

(cos(x)sin(x))’ = cos(x) sin(x)’ + cos(x)’ sin(x)

• sin(x)’ = cos(x)
• cos(x)’ = −sin(x)

So we can substitute:

(cos(x)sin(x))’ = cos(x) cos(x) + −sin(x) sin(x)

Which simplifies to:

(cos(x)sin(x))’ = cos 2 (x) − sin 2 (x)

Why Does It Work?

When we multiply two functions f(x) and g(x) the result is the area fg :

The derivative is the rate of change, and when x changes a little then both f and g will also change a little (by Δf and Δg). In this example they both increase making the area bigger.

How much bigger?

Increase in area = Δ(fg) = fΔg + ΔfΔg + gΔf

As the change in x heads towards zero, the "ΔfΔg" term also heads to zero, and we get:

Alternative Notation

An alternative way of writing it (called Leibniz Notation) is:

d dx (uv) = u dv dx + v du dx

Here is our example from before in Leibniz Notation:

Becomes this:

d dx (cos(x)sin(x)) = cos(x) d(sin(x)) dx + sin(x) d(cos(x)) dx

• d dx sin(x) = cos(x)
• d dx cos(x) = −sin(x)

d dx (cos(x)sin(x)) = cos(x) cos(x) + −sin(x) sin(x)

d dx (cos(x)sin(x)) = cos 2 (x) − sin 2 (x)

Three Functions

For three functions multiplied together we can use:

(fgh)’ = f’gh + fg’h + fgh’

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25 Product Rule

Okay, but what about [latex]\frac{d}{dx} x \cdot e^x[/latex]? Can we just take the derivative of each like this? \begin{align*} \frac{d}{dx} x \cdot e^x & = \left( \frac{d}{dx} x \right) \cdot \left( \frac{d}{dx} e^x \right) \\ & = 1 \cdot e^x = e^x \end{align*}

Unfortunately, no. Just to be clear: the above calculation is false!

Think about [latex]\frac{d}{dx} x \cdot x[/latex]. We know [latex]\frac{d}{dx} x = 1[/latex], so therefore you might think

[latex]\frac{d}{dx} x \cdot x = 1 \cdot 1 = 1[/latex]

But we also know that [latex]\frac{d}{dx} x \cdot x = \frac{d}{dx} x^2 = 2x[/latex]. So what do we make of this? Well, we just have to give up on the idea of “taking the derivative of each” with products. Luckily, there is a rule called the product rule that works great:

[latex]\boxed{\cfrac{d}{dx} f \cdot g = f g' + g f'}[/latex]

The motto for this rule is “the first times the derivative of the second, plus the second times the derivative of the first.”

Where does this rule come from? Well, consider this picture:

Here we have a rectangle. The height is [latex]f[/latex], the width is [latex]g[/latex]. The area of the rectangle is [latex]f \cdot g[/latex]. If we were to make [latex]f[/latex] bigger by a little bit, by say [latex]\Delta f[/latex], and [latex]g[/latex] bigger by [latex]\Delta g[/latex], then the rectangle would become bigger too. We want to find out how quickly the area is increasing. So how quickly is it increasing? You can see in the picture we add three sections on: [latex]f \cdot \Delta g[/latex], [latex]g \cdot \Delta f[/latex], and [latex]\Delta f \cdot \Delta g[/latex]. If you think of [latex]\Delta f[/latex] and [latex]\Delta g[/latex] as being really small, though, the [latex]\Delta f \cdot \Delta g[/latex] terms is incredibly tiny — so tiny it does not affect the answer in the limit. Therefore, the new area is [latex]f \cdot \Delta g + g \cdot \Delta f[/latex]. This is where the product rule comes from — it’s how an area changes as you change each side length.

Let’s see this rule in action.

Product Rule with [latex]x e^x[/latex]

In this case, we can attack this using the product rule with [latex]{\color{blue} f = x}[/latex], and [latex]{\color{red} g = e^x}[/latex]. We can easily take the derivative of each part: [latex]{\color{blue} f' = 1}[/latex], and [latex]{\color{red} g' = e^x}[/latex]. Hence, using the formula [latex]\frac{d}{dx} {\color{blue} f} {\color{red} g} = {\color{blue} f} {\color{red} g'} + {\color{red} g} {\color{blue} f'}[/latex], we have

[latex]{\color{blue} f} {\color{red} g'} + {\color{red} g} {\color{blue} f'} = {\color{blue} x} {\color{red} e^x} + {\color{red} e^x} {\color{blue} (1)} = \boxed{x e^x + e^x}.[/latex]

Another Product Rule

We see [latex]{\color{red} f = x^2, f' = 2x}[/latex] and [latex]{\color{blue} g = \cos(x), g' = -\sin(x)}[/latex]. Hence, we have \begin{align*} \frac{d}{dx} {\color{red} x^2} {\color{blue} \cos(x)} & = {\color{red} f} {\color{blue} g’} + {\color{blue} g} {\color{red} f’} \\ & = {\color{red} x^2} {\color{blue} (-\sin(x))} + {\color{blue} \cos(x)} {\color{red} 2x} \\ & = \boxed{-2x^2 \sin(x) + 2x \cos(x)}. \end{align*}

Product Rule with [latex]x \cdot x[/latex]

Using the power rule, we see [latex]\frac{d}{dx} x \cdot x = \frac{d}{dx} x^2 = 2x[/latex]. Using the product rule, we set [latex]f = x[/latex] and [latex]g = x[/latex]. In which case, [latex]f' = 1[/latex] and [latex]g' = 1[/latex]. Thus

[latex]f g' + g f' = (x)(1) + (x)(1) = \boxed{2x}.[/latex]

Hurray! The same answer! Isn’t it great when math just plain works.

We see [latex]f = \sqrt{x}[/latex], [latex]g = \ln(x)[/latex]. For [latex]f'[/latex], we rewrite [latex]f = x^{1/2}[/latex] and use the power rule to get [latex]f' = \frac{1}{2} x^{-1/2}[/latex], which is [latex]f' = \frac{1}{2 \sqrt{x}}[/latex]. As we saw in the exponents and logarithms section, [latex]g' = \frac{1}{x}[/latex]. Hence \begin{align*} \frac{d}{dx} \sqrt{x} \ln(x) & = f g’ + g f’ \\ & = \sqrt{x} \frac{1}{x} + \ln(x) \frac{1}{2 \sqrt{x}} \\ & = \frac{\sqrt{x}}{x} + \frac{\ln(x)}{2 \sqrt{x}} \end{align*} Now if you want to be a bit fancy with the algebra, we can simplify [latex]\frac{\sqrt{x}}{x}[/latex] using rational exponents. It is equal to [latex]\frac{x^{1/2}}{x^1}[/latex], which is [latex]x^{1/2-1} = x^{-1/2}[/latex]. If we turn this back into a root, we get [latex]\frac{1}{\sqrt{x}}[/latex]. From here, we can find a common denominator and add the two fractions.\begin{align*} \frac{d}{dx} \sqrt{x} \ln(x) & = \frac{1}{\sqrt{x}} + \frac{\ln(x)}{2 \sqrt{x}} \\ & = \frac{2}{2 \sqrt{x}} + \frac{\ln(x)}{2 \sqrt{x}} \\ & = \boxed{\frac{\ln(x) + 2}{2 \sqrt{x}}}. \end{align*} There you go.

Informal Calculus Copyright © by Tyler Seacrest is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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• Proof of the Product Rule

Key Questions

All we need to do is use the definition of the derivative alongside a simple algebraic trick.

First, recall the the the product #fg# of the functions #f# and #g# is defined as #(fg)(x)=f(x)g(x)# . Therefore, it's derivative is

#(fg)^(prime)(x) = lim_(h to 0) ((fg)(x+h)-(fg)(x))/(h) = lim_(h to 0) (f(x+h)g(x+h)-f(x)g(x))/(h)#

Now, note that the expression above is the same as

#lim_(h to 0) (f(x+h)g(x+h)+0-f(x)g(x))/(h)#

Wich we can rewrite, taking into account that #f(x+h)g(x)-f(x+h)g(x)=0# , as:

#lim_(h to 0) 1/h [f(x+h)g(x+h)+(f(x+h)g(x)-f(x+h)g(x))-f(x)g(x)] = lim_(h to 0) 1/h(f(x+h)[g(x+h)-g(x)]+g(x)[f(x+h)-f(x)])#

Using the property that the limit of a sum is the sum of the limits, we get:

#lim_(h to 0) f(x+h)(g(x+h)-g(x))/(h) + lim_(h to 0)g(x)(f(x+h)-f(x))/(h)#

Wich give us the product rule

#(fg)^(prime)(x) = f(x)g^(prime)(x)+g(x)f^(prime)(x),#

#lim_(h to 0) f(x+h) = f(x),# #lim_(h to 0)(g(x+h)-g(x))/(h) = g^(prime)(x),# #lim_(h to 0) g(x)=g(x),# #lim_(h to 0) (f(x+h)-f(x))/(h) = f^(prime)(x)#

Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher)

Good question, and that’s what we’re covering in today’s calculus class.

Let’s jump to it!

The Product Rule

Simply put, the term “product” means two functions are being multiplied together.

Discovered by Gottfried Leibniz, this rule allows us to calculate derivatives that we don’t want (or can’t) multiply quickly.

In other words, the product rule allows us to find the derivative of two differentiable functions that are being multiplied together by combining our knowledge of both the power rule and the sum and difference rule for derivatives.

Product Rule Formula

To better understand this formula, let’s work through a few examples, using color-coding to make things easier to follow.

Worked Example

For instance, let’s find the following derivative.

Derivative Product Rule Binomial

See, using the product rule is pretty easy!

Common Mistakes Using Product Rule

But I do want to caution you — the derivative of a product does not equal the product of derivatives.

Using the same two functions from above, let’s see this cautionary tale unfold:

Product Of Two Derivatives

But as we can see, the derivative of the product is not the same as the product of derivatives as:

Derivative Of Product Does NOT Equal Product Of Derivative

This is a very common mistake that students make, but now that you know what “not to do,” you’re well on your way to success.

Ex) Instantaneous Rate Of Change

Okay, so let’s investigate two more examples.

Suppose we are asked to find the derivative of the h(x) when x = 1, as seen below.

First, we will need to use the product rule to calculate the derivative, and then we will plug in the value of 1 to find the instantaneous rate of change.

Use Product Rule To Find The Instantaneous Rate Of Change

So, all we did was rewrite the first function and multiply it by the derivative of the second and then add the product of the second function and the derivative of the first. And lastly, we found the derivative at the point x = 1 to be 86.

Now for the two previous examples, we had options for how to find the derivative:

• Simplify first, then use the power rule to find the derivative
• Or use the product rule and then simplify

Ex) Trig Functions

So, will there ever be a time when the product rule is the only option for computing the derivative?

Yes! In the following example, we must use the product rule to find the rate of change because we cannot simplify the expression any further as we are dealing with a product of two functions entirely unrelated.

Product Rule Trig Functions

Note that we will learn how to find the derivative of trig functions in a future lesson. I merely wanted to demonstrate the “awesomeness” of this rule with this example and prove its usefulness.

Ex) More Than Two Functions

And did you know that we can extend this idea to the product of more than two functions?

For example, suppose we wanted to find the derivative of the product of three functions. You can use the following framework to accomplish this task.

Product Rule For Three Functions

And it also works with more functions in a similar manner.

Together we will walk through countless examples of using the product rule and applying our algebra skills to simplify our results for differentiating problems where one function is multiplied by another.

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Product rule

The product rule is a formula that is used to find the derivative of the product of two or more functions. Given two differentiable functions, f(x) and g(x), where f'(x) and g'(x) are their respective derivatives, the product rule can be stated as,

or using abbreviated notation:

The product rule can be expanded for more functions. For example, for the product of three functions, f, g, and h, the product rule is:

It can be expanded in this same way for as many products of functions as necessary. Below are some examples of use of the product rule for different cases.

Use the product rule to find the following derivatives.

1 . y = e x sin(x)

Let f(x) = e x and g(x) = sin(x), then apply the product rule:

2 . y = x 3 cos(x)ln(x)

Let f(x) = x 3 , g(x) = cos(x), and h(x) = ln(x), then apply the product rule:

Note that the product rule, like the quotient rule , chain rule , and others, is simply a method of differentiation. It can be used on its own, or in combination with other methods. The following examples will use the quotient rule and chain rule in addition to the product rule; refer to the quotient and chain rule pages for more information on the rules.

Differentiate the following functions.

In order to differentiate this, we need to use both the product and the quotient rules. Given two differentiable functions f(x) and g(x) where g(x) ≠ 0, the quotient rule can be written as:

In this case, f(x) = x 3 tan(x), and g(x) = e x . Noting that we must use the product rule to differentiate f(x), y can be differentiated using both product and quotient rules as follows:

To find the derivative, we need to use both the product rule and the chain rule ; the chain rule is used to differentiate sin(2x + 1).

Product Rule Explanation

It is not always necessary to compute derivatives directly from the definition. Several rules have been developed for finding the derivatives without having to use the definition directly. These rules simplify the process of differentiation. The Product Rule is a formula developed by Leibniz used to find the derivatives of products of functions.

The Product Rule is defined as the product of the first function and the derivative of the second function plus the product of the derivative of the first function and the second function :

The Formula for the Product Rule

Product Rule Example

Find f'(x) of

We can see that there is a product, so we can apply the product rule. First, we take the product of the first term and the derivative of the second term.

Second, we take the product of the derivative of the first term and the second term.

Then, we add them together to get our derivative.

Notice that if we multiplied them together at the start, the product would be 21x 5 . Taking the derivative after we multiplied it out would give us the same answer – 105x 4 . The product rule helps take the derivative of harder products of functions that require you use the rule instead of multiplying them together beforehand.

Let’s look at a harder example:

Differentiate:

We can see that we cannot multiply first and then take the derivative. We must use the product rule.

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The product rule is a fundamental concept in mathematics that describes how to differentiate the product of two or more functions. It is a crucial tool for analyzing and manipulating expressions involving exponents, polynomials, and logarithmic functions.

congrats on reading the definition of Product Rule . now let's actually learn it.

5 Must Know Facts For Your Next Test

• The product rule states that the derivative of the product of two functions is equal to the product of the first function's derivative and the second function, plus the product of the first function and the derivative of the second function.
• The product rule is essential for differentiating expressions involving exponents, as it allows for the simplification of complex derivatives.
• When multiplying polynomials, the product rule can be used to efficiently combine the terms and simplify the resulting expression.
• Rational exponents can be simplified using the product rule, which helps in evaluating and graphing logarithmic functions.
• The properties of logarithms, such as the product rule, are crucial for manipulating and evaluating logarithmic expressions.

Review Questions

• The product rule is essential when working with expressions involving exponents. It allows you to differentiate the product of two functions, which is crucial for simplifying complex expressions with exponents. For example, when dealing with scientific notation, the product rule can be used to differentiate and manipulate expressions that involve the multiplication of numbers with different exponents. This helps in performing calculations and converting between different forms of scientific notation.
• When multiplying polynomials, the product rule can be used to efficiently combine the terms and simplify the resulting expression. The product rule states that the derivative of the product of two functions is equal to the product of the first function's derivative and the second function, plus the product of the first function and the derivative of the second function. This principle can be applied to the multiplication of polynomials, allowing you to break down the process and arrive at the final simplified expression.
• The product rule is crucial for simplifying expressions involving rational exponents and evaluating logarithmic functions. Rational exponents can be expressed as fractions, and the product rule can be used to manipulate these expressions and arrive at simplified forms. Additionally, the properties of logarithms, such as the product rule, are essential for transforming and evaluating logarithmic functions. By understanding and applying the product rule in these contexts, you can effectively work with a wide range of mathematical expressions and functions.

Related terms

Exponents : Exponents represent the number of times a base number is multiplied by itself. They are used to express repeated multiplication concisely.

Polynomials : Polynomials are algebraic expressions consisting of variables and coefficients, where the variables are raised to non-negative integer powers.

Rational Exponents : Rational exponents are exponents that can be expressed as a fraction, allowing for the representation of fractional and negative powers.

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Product rule – Derivation, Explanation, and Example

The product rule allows us to find the derivative of two functions’ product using the respective functions’ corresponding derivatives.

This article will show how we can easily find the derivative of $f(x) \cdot g(x)$ and learn when this rule is most helpful. To master the product rule’s formula, we will derive the formula and try out a wide range of examples to show you how we can apply this product rule easily.

Finding the derivative of complex functions will also require other fundamental derivative rules, so make sure to refresh your notes on the previous rules you’ve already learned. For now, let’s dive right into understanding how the product rule works.

What is t he product rule ?                

The product rule is an essential derivative rule used to find the derivative of a function that can be expressed as a product of two simpler expressions . A great example of this type of function is $h(x) = (x^3 – 2x + 1)(x^3 – 4x^2 + 1)$.

Without the product rule, our option is to either use the formal definition of derivatives or expand the function to apply sum, difference, and power rules. But either way, it will be extremely tedious and will take a lot of our time. This is where the product rule enters.

According to the product rule, when we have $f(x)\cdot g(x)$, the derivative of this expression is simply the sum of the products of $\boldsymbol{f(x)}$ and $\boldsymbol{g’(x)}$ plus the product of $\boldsymbol{g(x)}$ and $\boldsymbol{f’(x)}$.

\begin{aligned} \dfrac{d}{dx} [f(x)\cdot g(x)] &= f(x) \dfrac{d}{dx} [g(x)]+ g(x) \dfrac{d}{dx} [f(x)]\\&= f*(x)g'(x) + g(x)f'(x)\end{aligned}

Through the product rule, we’ll only be needing the derivative of each factored expression and the original expressions that will be multiplied. Will this make the process easier? Definitely.

For some, since $f’(x)$ and $g’(x)$ will have lower degrees, the products $f’(x)g(x)$ and $g’(x)f(x)$ will be much simpler.

Deriving the product rule formula

Now that we’ve understood how this rule works let’s appreciate it even more by understanding how this product rule was derived using limit laws and the derivative’s formal definition.

If we have $f(x) \cdot g(x)$, we can find $\dfrac{d}{dx} [f(x)\cdot g(x)]$ using the formal definition of derivatives.

\begin{aligned} \dfrac{d}{dx} [f(x)\cdot g(x)] &= \lim_{h \rightarrow} \dfrac{[f(x +h)\cdot g(x+h)] – [f(x)\cdot g(x)]}{h}\end{aligned}

We can rewrite this expression using different algebraic manipulation techniques, as shown below.

\begin{aligned} \dfrac{d}{dx} [f(x)\cdot g(x)] &= \lim_{h \rightarrow 0} \dfrac{f(x+h)g(x +h)-f(x)g(x + h) + f(x)g(x + h) – f(x)g(x)}{h}\\&=\lim_{h \rightarrow 0} \left\{\left[\dfrac{f(x + h) -f(x)}{h} \right ]g(x + h) – f(x )\left[\dfrac{g(x + h) -g(x)}{h} \right ]\right\}\end{aligned}

We can use the limit laws and factor out $f(x)$ and $g(x)$ whenever possible.

\begin{aligned} \dfrac{d}{dx} [f(x)\cdot g(x)] &= \left[\lim_{h \rightarrow 0}\dfrac{f(x + h) -f(x)}{h} \right ] \lim_{h \rightarrow 0} g(x + h) + f(x )\left[\lim_{h \rightarrow 0}\dfrac{g(x + h) -g(x)}{h} \right ]\\&= \dfrac{d}{dx} [f(x)] \cdot g(x + 0) + [\dfrac{d}{dx} g(x)] \cdot f(x)\\&= f'(x)g(x) + g'(x)f(x)\end{aligned}

Here’s a visualization of how the product rule can be derived as well:

Hence, we’ve established the fact that $\dfrac{d}{dx} [f(x)g(x)] = f(x)g’(x) + g(x)f’(x)$.

How to use the product rule?

Let’s go ahead and understand how we can apply the product rule to find the derivative of a function that benefits from this rule. Here are some tips that can help you in applying this rule easily:

Take note of the derivative of the two expressions.

Alternate the pairings when you start multiplying: pair the first factor with the derivative of the second and do the same for the second factor with the first factor’s derivative.

Add the two resulting expressions to find the derivative of the original function.

Once we have the expressions for $f(x)$, $g(x)$, $f’(x)$ and $g’(x)$, let us share a quick trick to help you master this topic:

We can write the expressions this way so we can cross-multiply them and come up with the derivative of $f(x)g(x)$:

\begin{aligned}\dfrac{d}{dx} [f(x) \cdot g(x)] = {\color{green}f(x) g’(x) } + {\color{blue} g(x)f’(x)} \end{aligned}

Let’s say we want to find the derivative of $x^2 \sin x$. We can use the product rule as well as the following rules to find $\dfrac{d}{dx} (x^2 \sin x)$:

Derivative of sine: $\dfrac{d}{dx} \sin x =\cos x$

Power rule: $\dfrac{d}{dx} x^n = nx^{n-1}$

We can express $x^2 \sin x$ as a product of $x^2$ and $\sin x$, so this shows that we can use the product rule as shown below:

This shows that through the product rule, we can immediately find the derivative of $x^2 \sin x$. In fact, we have $\dfrac{d}{dx} (x^2\sin x) = x^2\cos x + 2x \sin x$.

Check out the sample problems below for you to work on and help you master this technique.

Find the derivative of $h(x) = 3x^5 \tan x$ using the product rule.

We can write $h(x)$ as a product of two expressions: $(3x^5)$ and $\tan x$. This shows that we can use the product rule to differentiate $h(x)$. Let’s recall the following derivative rules (aside from the product rule) that we’ll be using to find $h’(x)$:

The constant multiple rule: $\dfrac{d}{dx} cf(x) = c\dfrac{d}{dx} f(x)$.

The power rule: $\dfrac{d}{dx} x^n = nx^{n -1}$.

The derivative of tangent: $\dfrac{d}{dx} \tan x = \sec^2 x$.

This example shows you that thanks to the product rule, we can easily differentiate $h(x) = 3x^5 \tan x$. We have $h’(x) = 3x^5\sec^2 x + 15x^4 \tan x$ or $h’(x) = 3x^4(x \sec^2 x + 5\tan x)$.

Find the derivative of $y = e^x(x^6 – 5x^4 + 2x) $ using the product rule. (Here’s a tip: $\dfrac{d}{dx} e^x = e^x$)

We can see two factors here: $e^x$ and $x^6 – 5x^4 + 2x$. The tip already makes our lives easier since $\dfrac{d}{dx} e^x$ will remain $e^x$. For the second factor, we can use the following rules:

The sum rule for derivatives: $\dfrac{d}{dx} [f(x) + g(x)] = f’(x) + g(x)$.

Constant multiple rule: $\dfrac{d}{dx} cf(x) = c\dfrac{d}{dx} f(x)$.

Power rule: $\dfrac{d}{dx} x^n = nx^{n -1}$

Let’s go ahead and apply the product rule to find the derivative of $y=e^x(x^6 – 5x^4 + 2x) $.

This means that through product rule (and other fundamental derivative rules, of course), we were able to differentiate $y = e^x(x^6 – 5x^4 + 2x) $. Hence, we have $y’ =e^x(x^6 -5x^4 – 20x^3 + 2x +2)$.

Practice Questions

1. Find the derivative of $h(x) = -6x^4 \cos x$ using the product rule.

2. Find the derivative of $h(x) = 12x^8 \tan x$ using the product rule.

3.Find the derivative of $y = e^x \sin x$ using the product rule.

4.Find the derivative of $y = e^x (-12x^5 – 6x^2 + 7)$ using the product rule.

4.Find the derivative of $y = \sqrt{x}(x^4 – 1)$ using the product rule.

1. $h’(x) = -24x^3\cos x +6x^4 \sin x = 6x^3(-4\cos x + x\sin x)$

2. $h’(x) = 12x^8 \sec^2 x + 96x^7 \tan x = 12x^7(x\sec^2 x + 8\tan x)$

3. $y’ = e^x (\sin x + \cos x)$

4. $y’ = e^x (7 – 12 x – 6 x^2 – 60 x^4 – 12 x^5)$

5. $y’ = \dfrac{9}{2} x^{\frac{7}{2}} – \dfrac{1}{2\sqrt{x}} = \dfrac{9x^4 -1}{2\sqrt{x}}$

Images/mathematical drawings are created with GeoGebra.

Module 7: Exponents

Product and quotient rules, learning outcomes.

• Simplify exponential expressions with like bases using the product, quotient, and power rules

Use the Product Rule to Multiply Exponential Expressions

Exponential notation was developed to write repeated multiplication more efficiently. There are times when it is easier or faster to leave the expressions in exponential notation when multiplying or dividing. Let us look at rules that will allow you to do this.

For example, the notation [latex]5^{4}[/latex] can be expanded and written as [latex]5\cdot5\cdot5\cdot5[/latex], or [latex]625[/latex]. Do not forget, the exponent only applies to the number immediately to its left unless there are parentheses.

What happens if you multiply two numbers in exponential form with the same base? Consider the expression [latex]{2}^{3}{2}^{4}[/latex]. Expanding each exponent, this can be rewritten as [latex]\left(2\cdot2\cdot2\right)\left(2\cdot2\cdot2\cdot2\right)[/latex] or [latex]2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2[/latex]. In exponential form, you would write the product as [latex]2^{7}[/latex]. Notice that [latex]7[/latex] is the sum of the original two exponents, [latex]3[/latex] and [latex]4[/latex].

What about [latex]{x}^{2}{x}^{6}[/latex]? This can be written as [latex]\left(x\cdot{x}\right)\left(x\cdot{x}\cdot{x}\cdot{x}\cdot{x}\cdot{x}\right)=x\cdot{x}\cdot{x}\cdot{x}\cdot{x}\cdot{x}\cdot{x}\cdot{x}[/latex] or [latex]x^{8}[/latex]. And, once again, 8 is the sum of the original two exponents. This concept can be generalized in the following way:

The Product Rule for Exponents

For any number x and any integers a and b , [latex]\left(x^{a}\right)\left(x^{b}\right) = x^{a+b}[/latex].

To multiply exponential terms with the same base, add the exponents.

Write each of the following products with a single base. Do not simplify further.

• [latex]{t}^{5}\cdot {t}^{3}[/latex]
• [latex]\left(-3\right)^{5}\cdot \left(-3\right)[/latex]
• [latex]{x}^{2}\cdot {x}^{5}\cdot {x}^{3}[/latex]

Use the product rule to simplify each expression.

• [latex]{t}^{5}\cdot {t}^{3}={t}^{5+3}={t}^{8}[/latex]
• [latex]{\left(-3\right)}^{5}\cdot \left(-3\right)={\left(-3\right)}^{5}\cdot {\left(-3\right)}^{1}={\left(-3\right)}^{5+1}={\left(-3\right)}^{6}[/latex]

At first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first two.

Notice we get the same result by adding the three exponents in one step.

In the following video, you will see more examples of using the product rule for exponents to simplify expressions.

In our last product rule example, we will show that an exponent can be an algebraic expression. We can use the product rule for exponents no matter what the exponent looks like, as long as the base is the same.

Multiply. [latex]x^{a+2}\cdot{x^{3a-9}}[/latex]

We have two exponential terms with the same base. The product rule for exponents says that we can add the exponents.

[latex]x^{a+2}\cdot{x^{3a-9}}=x^{(a+2)+(3a-9)}=x^{4a-7}[/latex]

The expression cannot be simplified any further.

Use the Quotient Rule to Divide Exponential Expressions

Let us look at dividing terms containing exponential expressions. What happens if you divide two numbers in exponential form with the same base? Consider the following expression.

[latex] \displaystyle \frac{{{4}^{5}}}{{{4}^{2}}}[/latex]

You can rewrite the expression as: [latex] \displaystyle \frac{4\cdot 4\cdot 4\cdot 4\cdot 4}{4\cdot 4}[/latex]. Then you can cancel the common factors of 4 in the numerator and denominator: [latex] \displaystyle [/latex]

Finally, this expression can be rewritten as [latex]4^{3}[/latex] using exponential notation. Notice that the exponent, [latex]3[/latex], is the difference between the two exponents in the original expression, [latex]5[/latex] and [latex]2[/latex].

So, [latex] \displaystyle \frac{{{4}^{5}}}{{{4}^{2}}}=4^{5-2}=4^{3}[/latex].

Be careful that you subtract the exponent in the denominator from the exponent in the numerator.

So, to divide two exponential terms with the same base, subtract the exponents.

The Quotient (Division) Rule for Exponents

For any non-zero number x and any integers a and b : [latex] \displaystyle \frac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}[/latex]

• [latex]\dfrac{{\left(-2\right)}^{14}}{{\left(-2\right)}^{9}}[/latex]
• [latex]\dfrac{{t}^{23}}{{t}^{15}}[/latex]
• [latex]\dfrac{{\left(z\sqrt{2}\right)}^{5}}{z\sqrt{2}}[/latex]

Use the quotient rule to simplify each expression.

• [latex]\dfrac{{\left(-2\right)}^{14}}{{\left(-2\right)}^{9}}={\left(-2\right)}^{14 - 9}={\left(-2\right)}^{5}[/latex]
• [latex]\dfrac{{t}^{23}}{{t}^{15}}={t}^{23 - 15}={t}^{8}[/latex]
• [latex]\dfrac{{\left(z\sqrt{2}\right)}^{5}}{z\sqrt{2}}={\left(z\sqrt{2}\right)}^{5 - 1}={\left(z\sqrt{2}\right)}^{4}[/latex]

As we showed with the product rule, you may be given a quotient with an exponent that is an algebraic expression to simplify.  As long as the bases agree, you may use the quotient rule for exponents.

Simplify. [latex]\dfrac{y^{x-3}}{y^{9-x}}[/latex]

We have a quotient whose terms have the same base so we can use the quotient rule for exponents.

[latex]\dfrac{y^{x-3}}{y^{9-x}}=y^{(x-3)-(9-x)}=y^{2x-12}[/latex]

In the following video, you will see more examples of using the quotient rule for exponents.

• Simplify Expressions Using the Quotient Rule of Exponents (Basic). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/xy6WW7y_GcU . License : CC BY: Attribution
• College Algebra. Authored by : Abramson, Jay, et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]:1/Preface
• Ex: Expanding and Evaluating Exponential Notation . Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/KOnQpKSpVRo . License : CC BY: Attribution
• Ex: Simplify Exponential Expressions Using the Product Property of Exponents . Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/hA9AT7QsXWo . License : CC BY: Attribution

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Product rule formula and function notation

There is a formula that you can follow to use the product rule. It is important to try and remember this formula as it is not usually in exam formula booklets. If y = u v when u and v are functions of x then the product rule formula is:

\[\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\]

This can also be written in function notation ,

If \(f(x) = g(x)h(x)\) then \(f'(x) = g(x)h'(x) + h(x)g'(x)\)

Examples of the product rule

To understand the product rule further, let's look through some examples of using it.

If \(y = 5xe^2\) find \(\frac{dy}{dx}\):

First, you can start by looking at your product rule formula and finding each aspect of the formula:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v\frac{du}{dx}\]

If \(y = 5xe^2\), \(u = 5x\) and \(v = e^2\)

To find \(\frac{dv}{dx}\) and \(\frac{du}{dx}\) you can differentiate your u and v:

\(\frac{du}{dx} = 5\) \(\frac{dv}{dx} = 0\)

\[\frac{dy}{dx} = (5x)(0) + (e^2)(5) = 5e^2\]

You may also be asked to differentiate a trigonometric function using the product rule.

If \(y = (4\sin{x})e^2\) find \(\frac{dy}{dx}\)

As before, you can start by finding each part of the formula:

If \(y = (4\sin{x})e^2\), \(u = 4\sin x\) and \(v = e^2\)

Now you can differentiate your u and v to find \(\frac{dv}{dx}\) and \(\frac{du}{dx}\):

\(\frac{du}{dx} = 4\cos x\) \(\frac{dv}{dx} = 0\)

Finally, you can substitute each part into your formula to find \(\frac{dy}{dx}\):

\[\frac{dy}{dx} = (4\sin x)(0) + (e^2)(4\cos x)\]

\[\frac{dy}{dx} = 4e^2\cos x\]

You may even be asked to use the product rule in function notation , so let's work through an example.

If \(f(x) = 2x^2(x^2 + 4)\) find \(f'(x)\)

Once again, you can start by breaking down the formula for the product rule in function notation and finding each part.

\(f'(x) = g(x)h'(x) + h(x)g'(x)\)

If \(f(x) = 2x^2(x^2 + 4)\), \(g(x) = 2x^2\) and \(h(x) = x^2 + 4\)

Next, you can differentiate g(x) and h(x) to find the derivatives, g'(x) and h'(x):

\(g' = 4x\) \(h'(x) = 2x\)

Now that you have each part of the formula, you can solve it to find f'(x):

\(f'(x) = (2x^2)(2x) + (x^2 + 4)(4x) = 2(4x^3 + 8x)\)

If \(y = \ln x(x^2)\) find \(\frac{dy}{dx}\)

To begin with, let's look at the product rule formula and find each aspect of it:

If \(y = \ln x(x^2)\), \(u = \ln x\) and \(v = x^2\)

Let's differentiate the terms to find \(\frac{dv}{dx}\) and \(\frac{du}{dx}\):

\(\frac{du}{dx} = \frac{1}{x}\) \(\frac{dv}{dx} = 2x\)

Now you can put each part into the formula to find \(\frac{dy}{dx}\):

\[\frac{dy}{dx} = (\ln{x})(2x) + (x^2)(\frac{1}{x}) = 2x \ln{x} + x\]

Product Rule - Key takeaways

The product rule is one of the differentiation rules .

The product rule can be used when differentiating the products of two functions.

When using the product rule, you can either use the formula in y form or in the function notation form.

You may also need to differentiate trigonometric functions using the product rule.

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Frequently Asked Questions about Product Rule

What is the product rule?

The product rule is one of the differentiation rules that is used when you are differentiating the products of two functions.

When do you use the product rule or the chain rule?

You use the product rule when you are dealing with the products of two functions, and you use the chain rule when you are differentiating a composite function, also known as a function of a function.

How do you derive the product rule?

The product rule can be derived from the definition of a derivative,  dy/dx f(x) = lim (f(x+h)-f(x))/h

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Power of a Product Rule

Power of a Product: Algebra’s foundational principle of the “Power of a Product” makes expressions easier to understand and facilitates equation solving. This rule finds applications in a number of disciplines, including economics, computer science, and engineering. The purpose of this article is to describe the Power of a Product rule, examine its significance, and emphasize its characteristics and uses.

Table of Content

What is the Power of a Product?

• Power of a Product: Importance
• Power of a Product: Properties
• Power of a Product: Applications

Differences Between the Power Rule and Product Rule

• Power of a Product: Solved Examples
• Power of a Product: Practice Problems

Power of a Product: Common Mistakes

• Incorrect Distribution of Exponent:
• Confusing with Addition:
• FAQs on Power of a Product

The Power of a Product rule is an exponent property that states:

(ab) n = a n . b n

This means that when you raise a product of two or more numbers to a power, the exponent can be applied individually to each factor.

Properties of Product of Power Rule

Some of the common properties of product rule of powers are:

• Distributive Nature : The exponent distributes over the factors inside the parentheses.
• Associativity : The order in which the factors are grouped does not affect the outcome.
• Commutativity: The order of the factors does not matter.

Applications of Power of a Product

Simplifying Algebraic Expressions:

• Example: Simplify (2x) 3
• Solution: (2x) 3 = 2 3 ⋅ x 3 = 8x 3

Solving Equations

• Example : Solve (xy) 3 = 36.
• Solution : Apply the Power of a Product rule: x 2 ⋅ y 2 = 36. Then, solve for x and y based on additional information.

Calculating Compound Interest

Let’s consider an example for calculation of compound interest.

Example: Calculate the amount after 2 years with a principal of $1000 and an annual interest rate of 5% compounded yearly.

Use the formula A = P(1+r/n) nt Substituting the values, you get A=1000(1.05) 2 =1102.5

Power of a Product applies when raising a product to a power: (ab) n = a n ⋅ b n .

Product Rule in calculus refers to the differentiation of the product of two functions: (fg)′ = f′g + fg′

Power of a Product Rule: Solved Examples

Example 1: Simplify (3y 2 ) 4

Solution: (3y 2 ) 4 = 3 4 ⋅ (y 2 ) 4 = 81y 8

Example 2: Solve (2a 3 b) 2 =64a 6 b 2

Solution: Apply the Power of a Product rule and simplify.

Power of a Product Rule: Practice Problems

Simplify (5x 2 y) 3

Solve (mn) 3 = 27m 3 n 3

Calculate the final amount with a principal of $500 at an interest rate of 4% compounded annually for 3 years.

Some of the common mistakes while using Power of a Product Rule are:

Incorrect Distribution of Exponent

• Incorrect: (2x) 2 = 2x 2
• Correct: (2x) 2 = 2 2 . x 2 = 4 x 2

Confusing with Addition

• Incorrect: (x + y) 2 = x 2 + y 2
• Correct: (x + y) 2 = x 2 + 2xy + y 2
• Simplifying Exponents
• Laws of Exponent
• Product Rule of Derivative

FAQs on Power of a Product Rule

What is the power of a product rule.

The Power of a Product rule states that (ab) n = a n . b n

Why is the Power of a Product rule important in algebra?

It simplifies complex expressions and is essential for solving equations involving exponents.

Can the Power of a Product rule be applied to more than two factors?

Yes, it can be applied to any number of factors within the product.

What is the difference between the Power Rule and the Product Rule?

The Power Rule deals with raising a product to a power, while the Product Rule is used in differentiation.

How does the Power of a Product rule apply to compound interest calculations?

It is used in the formula A = P(1 + r/n) nt to calculate the total amount with compound interest.

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Decomposing tensor product of lie algebra representations

I'm given a lie algebra representation $\pi$ of some semi-simple algebra and that it decomposes into a sum of irreducible representations.

What technique should I use to show the decomposition of $\pi \otimes \pi$ into irreducible represenations?

Any clues will be highly appreciated.

Thanks in advance.

• representation-theory
• lie-algebras

• $\begingroup$ Do you just need to show that it decomposes, or do you need to find the summands and their multiplicities? For the latter, you can reduce it to figuring out how the tensor product of two arbitrary simple modules splits (which can actually be fairly complicated). $\endgroup$ –  Tobias Kildetoft Commented Apr 16, 2013 at 14:03
• $\begingroup$ I think I need to show the later. If it's the standard representation we are talking about, is it any easier? $\endgroup$ –  user71158 Commented Apr 16, 2013 at 15:01
• $\begingroup$ Well, the tensor product of the standard representation with itself has two submodules that are "easy" to find, namely the second symmetric power and the second exterior power. $\endgroup$ –  Tobias Kildetoft Commented Apr 16, 2013 at 17:22
• $\begingroup$ Right! Somehow I missed that one. But then, what is the significance of the fact that $\pi$ decomposes into sum of irreducible representations? Is it some precondition? $\endgroup$ –  user71158 Commented Apr 16, 2013 at 18:26
• $\begingroup$ Any finite dimensional representation of a semisimple Lie algebra decomposes into a direct sum of irreducibles, so that is not really extra information. $\endgroup$ –  Tobias Kildetoft Commented Apr 16, 2013 at 18:28

2 Answers 2

One approach to the general problem of decomposing a tensor product of irreducible finite-dimensional representations (hence any finite-dimensional representations) into irreducibles is to use the theory of crystals. The crystal of a representation is a colored directed graph associated to that representation. There is a purely combinatorial algorithm for producing the tensor product of two crystals. Then the connected components of the crystal graph correspond to the irreducible representations you're looking for. For many simple Lie algebras, the crystals of the irreducible finite-dimensional representations are described very explicitly. For instance, for $\mathfrak{sl}_n$, they are in terms of semistandard tableaux. So finding the decomposition you seek becomes combinatorics.

Depending on the sizes of the representations involved, you may be able to solve your decomposition problem using the Weyl character formula. This is especially true if your representations are finite-dimensional. If so, then the character is a complete invariant of the representation, and it is additive and multiplicative under direct sum and tensor product.

The Weyl character formula tells you the character of an irrep in terms of its highest weight. Conversely, given any character, you can find the highest weight with nonzero coefficient; subtract off the corresponding character, and repeat. In practice, this is pretty fast by hand for $\mathrm{sl}(3)$, and on a computer for larger things.

For more complicated problems, I've heard very good things about the computer algebra package LiE , although I have never used it myself.

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