Topics You Need To Understand For This Page
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 filter them from the list of practice problems. 
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. 
Calculus Main Topics
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Product Rule 
The Product Rule is pretty straightforward. 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 

practice filters 

use basic derivatives and product rule only (11) 

use trig rules (4) 

use exponential and/or logarithmic rules (4) 
hidden practice problems 
Instructions   Unless otherwise instructed, calculate the derivatives of the following functions using the product rule, giving your final answers in simplified, factored form.
Practice A01 
\(f(x)=x(25x)\) 

Practice A02 
\(f(x)=(x2)(x+3)\) 


\(f'(x)=2x+1\) 
Practice A02 Final Answer 
\(f'(x)=2x+1\) 
Practice A03 
\(f(x)=(2x+3)(3x2)\) 

Practice A04 
\(h(x)=(x^2+5x+7)(x^3+2x4)\) 


\(h'(x)=5x^4+20x^3+27x^2+12x6\) 
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. 
\( (x^2 + 5x + 7)(3x^2 + 2) + (x^3 + 2x  4)(2x + 5) \) 
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. 
\( (x^2)(3x^2 + 2) + 5x(3x^2 + 2) + 7(3x^2 + 2) + (x^3)(2x+5) + 2x(2x+5)  4(2x+5) \) 
\(3x^4 + 2x^2 + 15x^3 + 10x + 21x^2 + 14 + 2x^4 + 5x^3 + 4x^2 + 10x  8x  20 \) 
Collect common terms. 
\( 5x^4 + 20x^3 + 27x^2 + 12x  6 \) 
Practice A04 Final Answer 
\(h'(x)=5x^4+20x^3+27x^2+12x6\) 
Practice A05 
\(g(x)=(x^37x^2+4)(3x^2+14)\) 


\(g'(x)=x(15x^384x^2+42x172)\) 
First, we apply the product rule. 
\( \displaystyle{ \frac{dg}{dx} = }\) \(\displaystyle{ (x^37x^2+4)\cdot }\) \(\displaystyle{ \frac{d}{dx}[3x^2+14]+ }\)
\(\displaystyle{ (3x^2+14)\cdot }\) \(\displaystyle{ \frac{d}{dx}[x^37x^2+4] }\)

Now, we evaluate the derivatives in the above equation. 
\( \displaystyle{ \frac{dg}{dx} = }\) \((x^37x^2+4)(6x)\) \(+\) \((3x^2+14)(3x^214x)\) 
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. 
\(x[6x^342x^2+24+(3x^2+14)(3x14)]\) 
Now, we multiply out the terms inside the brackets. 
\(x(15x^384x^2+42x172)\) 
Factoring right after taking the derivative, before multiplying out, makes it easier to see common terms.
Practice A05 Final Answer 
\(g'(x)=x(15x^384x^2+42x172)\) 
Practice A06 
\(g(x)=(3x^4+2x1)(x^52x^2)\) 

Practice A07 
\(\displaystyle{f(x)=x^3e^x}\) 


\(\displaystyle{\frac{d}{dx}\left[x^3e^x\right]=x^2e^x(x+3)}\) 
\(\displaystyle{ \frac{d}{dx}\left[x^3e^x\right] }\) 
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. 
\( x^2e^x(x+3) \) 
Practice A07 Final Answer 
\(\displaystyle{\frac{d}{dx}\left[x^3e^x\right]=x^2e^x(x+3)}\) 
Practice A08 
\(\displaystyle{y=7e^{2x}}\) 


\(\displaystyle{\frac{d}{dx}\left[7e^{2x}\right]=14e^{2x}}\) 
\(\displaystyle{ \frac{d}{dx} \left[ 7e^{2x} \right] }\) 
\(\displaystyle{ 7 \frac{d}{dx}[ e^x \cdot e^x ] }\) 
\(\displaystyle{ 7e^x \frac{d[e^x]}{dx} + 7e^x \frac{d[e^x]}{dx} }\) 
\( 7e^x (e^x) + 7e^x (e^x) \) 
\( 14e^{2x} \) 
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 if you do not know the chain rule yet.
Practice A08 Final Answer 
\(\displaystyle{\frac{d}{dx}\left[7e^{2x}\right]=14e^{2x}}\) 
Practice A09 
\(\displaystyle{e^x\sqrt{x}}\) 


\( \displaystyle{\frac{d}{dx}\left[ e^x \sqrt{x} \right] = \left[ \frac{e^x}{2x^{1/2}} \right] (1 + 2x)} \) 
\(\displaystyle{ \frac{d}{dx}\left[ e^x \sqrt{x} \right] }\) 
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.
Practice A09 Final Answer 
\( \displaystyle{\frac{d}{dx}\left[ e^x \sqrt{x} \right] = \left[ \frac{e^x}{2x^{1/2}} \right] (1 + 2x)} \) 
Practice A10 
\(\displaystyle{x^4\tan(x)}\) 


\(\displaystyle{\frac{d}{dx}[ x^4 \tan(x) ] = x^3[x \sec^2(x)+4 \tan(x)]}\) 
\(\displaystyle{\frac{d}{dx}[ x^4 \tan(x) ] = x^4 \sec^2(x) + 4x^3 \tan(x) = }\) \(\displaystyle{x^3[x \sec^2(x)+4 \tan(x)]}\)
Note: Not factoring out the \(x^3\) term, may cost you points.
Practice A10 Final Answer 
\(\displaystyle{\frac{d}{dx}[ x^4 \tan(x) ] = x^3[x \sec^2(x)+4 \tan(x)]}\) 
Practice A11 
\(\displaystyle{f(x)=\cos(x)\sin(x)}\) 


\(\displaystyle{ \frac{d}{dx}\left[ \cos(x) \sin(x) \right] = \cos^2(x)  \sin^2(x) }\) 
\(\displaystyle{ \frac{d}{dx}\left[ \cos(x) \sin(x) \right] }\) 
\(\displaystyle{ cos(x) \frac{d}{dx}[ \sin(x)] + \sin(x) \frac{d}{dx}[ \cos(x)] }\) 
\( \cos(x)[ \cos(x)] + \sin(x)[\sin(x)] \) 
\( \cos^2(x)  \sin^2(x) \) 
Practice A11 Final Answer 
\(\displaystyle{ \frac{d}{dx}\left[ \cos(x) \sin(x) \right] = \cos^2(x)  \sin^2(x) }\) 
Practice B01 
\(\displaystyle{g(x)=(x^2+3x5)\sqrt[3]{x}}\) 


\(\displaystyle{\frac{d}{dx}\left[(x^2+3x5)\sqrt[3]{x}\right]=\frac{7x^2+12x5}{3x^{2/3}}}\) 
Rewrite \(\sqrt[3]{x}\) as \( x^{1/3} \) to use the power rule. 
\(\displaystyle{ \frac{dg}{dx} = \frac{d}{dx}\left[ (x^2+3x5)(x^{1/3}) \right] }\) 
Use the product rule. 
\(\displaystyle{ (x^2 + 3x  5) \cdot \frac{d[x^{1/3}]}{dx} + \left( x^{1/3} \right) \cdot \frac{d[x^2+3x5]}{dx} }\) 
Take the two derivatives. 
\( (x^2 + 3x  5) \cdot (1/3)x^{2/3} + x^{1/3} \cdot (2x+3) \) 
\(\displaystyle{ \frac{ x^2 + 3x  5}{ 3x^{2/3}} + \left[ x^{1/3} \cdot (2x+3) \right] \left[\frac{3x^{2/3}}{3x^{2/3}} \right] }\) 
\(\displaystyle{ \frac{ x^2 + 3x  5 + (3x)(2x+3) }{3x^{2/3}} }\) 
\(\displaystyle{ \frac{7x^2 + 12x  5}{3x^{2/3}} }\) 
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 \(\sqrt[3]{x^2}\). 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.
Practice B01 Final Answer 
\(\displaystyle{\frac{d}{dx}\left[(x^2+3x5)\sqrt[3]{x}\right]=\frac{7x^2+12x5}{3x^{2/3}}}\) 
Practice B02 
\(\displaystyle{f(x)=\left[\frac{1}{x^2}\frac{3}{x^4}\right](x+5x^3)}\) 


\(5+14/x^2+9/x^4\) 
Both of these videos contain the solution to this problem. We are including both here in case one of them helps you more than the other.
Practice B02 Final Answer 
\(5+14/x^2+9/x^4\) 
Practice B03 
\(\displaystyle{H(u)=(u\sqrt{u})(u+\sqrt{u})}\) 

Practice B04 
\(f(x)=x(2x1)(2x+1)\) 

Practice B05 
\(f(x)=(2x+1)(4x^2)(1+x^2)\) 

Practice B06 
\(\displaystyle{\frac{3}{x}\cot(x)}\) 


\(\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] = \frac{3}{x^2}\left[ x \csc^2(x) + \cot(x) \right] }\) 
\(\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] }\) 
\(\displaystyle{ \frac{3}{x}\frac{d}{dx}[\cot(x)] + \cot(x) \frac{d}{dx}[3/x] }\) 
\(\displaystyle{ \frac{3}{x} (\csc^2(x)) + \cot(x) \frac{d}{dx}[3x^{1}] }\) 
\(\displaystyle{ \frac{3}{x} \csc^2(x) + \cot(x) [3(1)x^{2}] }\) 
\(\displaystyle{ \frac{3}{x} \csc^2(x)  \frac{3}{x^2} \cot(x) }\) 
\(\displaystyle{ \frac{3}{x^2}[ x \csc^2(x) + \cot(x) ] }\) 
Practice B06 Final Answer 
\(\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] = \frac{3}{x^2}\left[ x \csc^2(x) + \cot(x) \right] }\) 
Practice C01 
\(\displaystyle{f(x)=4x^2e^x\sin(x)\sec(x)}\) 
