## 17Calculus Derivatives - Product Rule

Product Rule

The Product Rule is pretty straight-forward. If you have a function with two main parts that are multiplied together, for example $$h(x)=f(x) \cdot g(x)$$, the derivative is

$$\displaystyle{ \frac{dh}{dx} = }$$ $$\displaystyle{ \frac{d}{dx}[f \cdot g] = }$$ $$\displaystyle{ f'\cdot g + f \cdot g' }$$

An interesting thing to notice about the product rule is that the constant multiple rule is just a special case of the product rule. For example, if you have function $$f(x) = cg(x)$$, the product rule says $$f'(x) = (c)' g(x) + c g'(x) =$$ $$0 + c g'(x) = c g'(x)$$. Notice that we use the constant rule to say that $$d[c]/dx = 0$$.

And that's all you need to know to use the product rule. A common mistake many students make is to think that the product rule allows you to take the derivative of both terms and multiply them together. WRONG! If this confuses you, go back to the top of the page and reread the product rule and then go through some examples in your textbook.

Before you start using the product rule, it is important to know where it comes from. So take a few minutes to watch this video showing the proof of the product rule.

### PatrickJMT - Product Rule Proof [6min-6secs]

video by PatrickJMT

 Okay, practice problem time. Once you are finished with those, the quotient rule is the next logical step.

### Practice

Instructions - - Unless otherwise instructed, calculate the derivatives of the following functions using the product rule, giving your final answers in simplified, factored form.

Basic Problems

Here are some problems that use only the product rule, the power rule and the other basic rules on the main derivatives page.

$$f(x) = x(25-x)$$

Problem Statement

Use the product rule to calculate the derivative of $$f(x) = x(25-x)$$

Solution

### 938 video

video by Krista King Math

$$f(x) = (x-2)(x+3)$$

Problem Statement

Use the product rule to calculate the derivative of $$f(x) = (x-2)(x+3)$$

$$f'(x)=2x+1$$

Problem Statement

Use the product rule to calculate the derivative of $$f(x) = (x-2)(x+3)$$

Solution

### 1928 video

video by PatrickJMT

$$f'(x)=2x+1$$

$$f(x) = (2x+3)(3x-2)$$

Problem Statement

Use the product rule to calculate the derivative of $$f(x) = (2x+3)(3x-2)$$

Solution

### 939 video

video by Krista King Math

$$h(x)=(x^2+5x+7)(x^3+2x-4)$$

Problem Statement

Use the product rule to calculate the derivative of $$h(x)=(x^2+5x+7)(x^3+2x-4)$$

$$h'(x)=5x^4+20x^3+27x^2+12x-6$$

Problem Statement

Use the product rule to calculate the derivative of $$h(x)=(x^2+5x+7)(x^3+2x-4)$$

Solution

 Use the product rule. $$\displaystyle{ h'(x) = (x^2 + 5x + 7)\frac{d}{dx}[x^3 + 2x - 4] + (x^3 + 2x - 4) \frac{d}{dx}[x^2 + 5x + 7] }$$ Take the derivative of the appropriate terms. $$h'(x) = (x^2 + 5x + 7)(3x^2 + 2) + (x^3 + 2x - 4)(2x + 5)$$ Check for common terms. In this case, there are not any, so multiply out. Leaving it in this form is not usually considered proper form but as usual check with your instructor to see what they expect. $$h'(x) = (x^2)(3x^2 + 2) + 5x(3x^2 + 2) + 7(3x^2 + 2) + (x^3)(2x+5) + 2x(2x+5) - 4(2x+5)$$ $$h'(x) = 3x^4 + 2x^2 + 15x^3 + 10x + 21x^2 + 14 + 2x^4 + 5x^3 + 4x^2 + 10x - 8x - 20$$ Collect common terms. $$h'(x) = 5x^4 + 20x^3 + 27x^2 + 12x - 6$$

$$h'(x)=5x^4+20x^3+27x^2+12x-6$$

$$g(x) = (x^3-7x^2+4)(3x^2+14)$$

Problem Statement

Use the product rule to calculate the derivative of $$g(x) = (x^3-7x^2+4)(3x^2+14)$$

$$g'(x) = x(15x^3-84x^2+42x-172)$$

Problem Statement

Use the product rule to calculate the derivative of $$g(x) = (x^3-7x^2+4)(3x^2+14)$$

Solution

 First, we apply the product rule. $$\displaystyle{ \frac{dg}{dx} = }$$ $$\displaystyle{ (x^3-7x^2+4)\cdot \frac{d}{dx}[3x^2+14]+ }$$ $$\displaystyle{ (3x^2+14)\cdot \frac{d}{dx}[x^3-7x^2+4] }$$ Now, we evaluate the derivatives in the above equation. $$\displaystyle{ \frac{dg}{dx} = }$$ $$(x^3-7x^2+4)(6x)$$ $$+$$ $$(3x^2+14)(3x^2-14x)$$ Before we simplify by multiplying out, let's look for common factors that we can factor out. Notice that there is an $$x$$ in both of the terms. So we factor that out. $$g'(x) = x[6x^3-42x^2+24+(3x^2+14)(3x-14)]$$ Now, we multiply out the terms inside the brackets. $$g'(x) = x(15x^3-84x^2+42x-172)$$

Factoring right after taking the derivative, before multiplying out, makes it easier to see common terms.

$$g'(x) = x(15x^3-84x^2+42x-172)$$

$$g(x) = (3x^4+2x-1)(x^5-2x^2)$$

Problem Statement

Use the product rule to calculate the derivative of $$g(x) = (3x^4+2x-1)(x^5-2x^2)$$

Solution

### 1311 video

video by PatrickJMT

These problems require you to know how to take the derivative of exponential functions.

$$\displaystyle{f(x)=x^3e^x}$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{f(x)=x^3e^x}$$

$$\displaystyle{\frac{d}{dx}\left[x^3e^x\right]=x^2e^x(x+3)}$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{f(x)=x^3e^x}$$

Solution

 $$\displaystyle{ \frac{d}{dx}\left[x^3e^x\right] }$$ Use the product rule. $$\displaystyle{ x^3\cdot\frac{d[e^x]}{dx}+e^x\cdot\frac{d[x^3]}{dx} }$$ Take the derivative of the two terms. $$x^3\cdot e^x+e^x\cdot3x^2$$ Simplify by factoring. $$x^2e^x(x+3)$$

$$\displaystyle{\frac{d}{dx}\left[x^3e^x\right]=x^2e^x(x+3)}$$

$$\displaystyle{y=7e^{2x}}$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{y=7e^{2x}}$$

$$\displaystyle{\frac{d}{dx}\left[7e^{2x}\right]=14e^{2x}}$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{y=7e^{2x}}$$

Solution

 $$\displaystyle{ \frac{d}{dx} \left[ 7e^{2x} \right] }$$ $$\displaystyle{ 7 \frac{d}{dx}[ e^x \cdot e^x ] }$$ $$\displaystyle{ 7e^x \frac{d[e^x]}{dx} + 7e^x \frac{d[e^x]}{dx} }$$ $$7e^x (e^x) + 7e^x (e^x)$$ $$14e^{2x}$$

Note: The derivative is more easily calculated using the chain rule. However, since you are not required to know the chain rule for this page, this is a valid way to work it without the chain rule.

$$\displaystyle{\frac{d}{dx}\left[7e^{2x}\right]=14e^{2x}}$$

$$\displaystyle{e^x\sqrt{x}}$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{e^x\sqrt{x}}$$

$$\displaystyle{ \frac{e^x (1 + 2x)}{2x^{1/2}} }$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{e^x\sqrt{x}}$$

Solution

 $$\displaystyle{ \frac{d}{dx}\left[ e^x \sqrt{x} \right] }$$ Rewrite $$\sqrt{x}$$ as $$x^{1/2}$$ so that we can use the power rule. $$\displaystyle{ \frac{d}{dx}\left[ e^x \left( x^{1/2} \right) \right] }$$ Use the product rule. $$\displaystyle{ (e^x) \frac{d[x^{1/2}]}{dx} + \left( x^{1/2} \right) \frac{d[e^x]}{dx} }$$ $$e^x (1/2)x^{-1/2} + x^{1/2} e^x$$ $$\displaystyle{ \left[ \frac{e^x}{2x^{1/2}} \right] (1 + 2x) }$$

Notes -
After we used the product rule, we just used algebra to simplify and factor. Although there are many ways to write the final answer, we usually want all factors written with positive exponents, except possibly exponential terms. Your instructor may or may not enforce this but most textbooks will write it this way. Writing your answer as $$e^x \left[ (1/2)x^{-1/2} + x^{1/2} \right]$$ is not considered completely factored because both factors inside the parentheses have an x term.

$$\displaystyle{ \frac{e^x (1 + 2x)}{2x^{1/2}} }$$

These problems require you to know how to take the derivative of trig functions.

$$\displaystyle{x^4\tan(x)}$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{x^4\tan(x)}$$

$$\displaystyle{ x^3[x \sec^2(x)+4 \tan(x)] }$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{x^4\tan(x)}$$

Solution

$$\displaystyle{\frac{d}{dx}[ x^4 \tan(x) ] = x^4 \sec^2(x) + 4x^3 \tan(x) = }$$ $$\displaystyle{x^3[x \sec^2(x)+4 \tan(x)]}$$

Note: Not factoring out the $$x^3$$ term, may cost you points. Check with your instructor to see what they expect.

$$\displaystyle{ x^3[x \sec^2(x)+4 \tan(x)] }$$

$$\displaystyle{f(x)=\cos(x)\sin(x)}$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{f(x)=\cos(x)\sin(x)}$$

$$f'(x) = \cos^2(x) - \sin^2(x)$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{f(x)=\cos(x)\sin(x)}$$

Solution

 $$\displaystyle{ f'(x) = \frac{d}{dx}\left[ \cos(x) \sin(x) \right] }$$ $$\displaystyle{ f'(x) = \cos(x) \frac{d}{dx}[ \sin(x)] + \sin(x) \frac{d}{dx}[ \cos(x)] }$$ $$f'(x) = \cos(x)[ \cos(x)] + \sin(x)[-\sin(x)]$$ $$f'(x) = \cos^2(x) - \sin^2(x)$$

$$f'(x) = \cos^2(x) - \sin^2(x)$$

Intermediate Problems

$$\displaystyle{g(x)=(x^2+3x-5)\sqrt{x}}$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{g(x)=(x^2+3x-5)\sqrt{x}}$$

$$\displaystyle{\frac{d}{dx}\left[(x^2+3x-5)\sqrt{x}\right]=\frac{7x^2+12x-5}{3x^{2/3}}}$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{g(x)=(x^2+3x-5)\sqrt{x}}$$

Solution

 Rewrite $$\sqrt{x}$$ as $$x^{1/3}$$ to use the power rule. $$\displaystyle{ g'(x) = \frac{d}{dx}\left[ (x^2+3x-5)(x^{1/3}) \right] }$$ Use the product rule. $$\displaystyle{ g'(x) = (x^2 + 3x - 5) \cdot \frac{d[x^{1/3}]}{dx} + \left( x^{1/3} \right) \cdot \frac{d[x^2+3x-5]}{dx} }$$ Take the two derivatives. $$g'(x) = (x^2 + 3x - 5) \cdot (1/3)x^{-2/3} + x^{1/3} \cdot (2x+3)$$ $$\displaystyle{ g'(x) = \frac{ x^2 + 3x - 5}{ 3x^{2/3}} + \left[ x^{1/3} \cdot (2x+3) \right] \left[\frac{3x^{2/3}}{3x^{2/3}} \right] }$$ $$\displaystyle{ g'(x) = \frac{ x^2 + 3x - 5 + (3x)(2x+3) }{3x^{2/3}} }$$ $$\displaystyle{ g'(x) = \frac{7x^2 + 12x - 5}{3x^{2/3}} }$$

Notes -
After we used the product rule, we just used algebra to simplify and factor. Although there are many ways to write the final answer, we usually want all factors written with positive exponents. Your instructor may or may not enforce this but most textbooks will write it this way.
Also, you may see $$x^{2/3}$$ written as $$\sqrt{x^2}$$. Although this is correct notation, in calculus we usually leave our answer in fractional exponent form. Ask your instructor which form he/she prefers or check your textbook to see which form it shows the most.

$$\displaystyle{\frac{d}{dx}\left[(x^2+3x-5)\sqrt{x}\right]=\frac{7x^2+12x-5}{3x^{2/3}}}$$

$$\displaystyle{f(x)=\left[\frac{1}{x^2}-\frac{3}{x^4}\right]\cdot}$$ $$(x+5x^3)$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{f(x)=\left[\frac{1}{x^2}-\frac{3}{x^4}\right]\cdot}$$ $$(x+5x^3)$$

Solution

### 935 video

video by Krista King Math

$$H(u)=(u-\sqrt{u})$$ $$(u+\sqrt{u})$$

Problem Statement

Use the product rule to calculate the derivative of $$H(u)=(u-\sqrt{u})$$ $$(u+\sqrt{u})$$

Solution

### 936 video

video by Krista King Math

$$f(x)=x(2x-1)(2x+1)$$

Problem Statement

Use the product rule to calculate the derivative of $$f(x)=x(2x-1)(2x+1)$$

Solution

### 940 video

video by Krista King Math

$$f(x)=(2x+1)$$ $$(4-x^2)(1+x^2)$$

Problem Statement

Use the product rule to calculate the derivative of $$f(x)=(2x+1)$$ $$(4-x^2)(1+x^2)$$

Solution

### 937 video

video by Krista King Math

Here is one with a trig derivative.

$$\displaystyle{\frac{3}{x}\cot(x)}$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{\frac{3}{x}\cot(x)}$$

$$\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] = \frac{-3}{x^2}\left[ x \csc^2(x) + \cot(x) \right] }$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{\frac{3}{x}\cot(x)}$$

Solution

 $$\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] }$$ $$\displaystyle{ \frac{3}{x}\frac{d}{dx}[\cot(x)] + \cot(x) \frac{d}{dx}[3/x] }$$ $$\displaystyle{ \frac{3}{x} (-\csc^2(x)) + \cot(x) \frac{d}{dx}[3x^{-1}] }$$ $$\displaystyle{ \frac{-3}{x} \csc^2(x) + \cot(x) [3(-1)x^{-2}] }$$ $$\displaystyle{ \frac{-3}{x} \csc^2(x) - \frac{3}{x^2} \cot(x) }$$ $$\displaystyle{ \frac{-3}{x^2}[ x \csc^2(x) + \cot(x) ] }$$

$$\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] = \frac{-3}{x^2}\left[ x \csc^2(x) + \cot(x) \right] }$$

Here is a problem that has both trig and exponential derivatives. Give it a try.

$$f(x)=4x^2e^x$$ $$\sin(x)\sec(x)$$

Problem Statement

Use the product rule to calculate the derivative of $$f(x)=4x^2e^x$$ $$\sin(x)\sec(x)$$

Solution

### 934 video

video by Krista King Math

You CAN Ace Calculus

 basic derivative rules power rule Some of the practice problems require you to know the following rules also (in their basic form, not including the chain rule). If you don't know one or more of these rules, no worries. You can skip those problems and come back to them later. exponential derivative derivatives of trig functions You do NOT need to know the chain rule for anything on this page, including practice problems. For practice problems using the product rule and chain rule, see the chain rule page.

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