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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Related Topics and Links
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 
\(h(x)=(x^2+5x+7)(x^3+2x4)\) 


\(h'(x)=5x^4+20x^3+27x^2+12x6\) 
Using the product rule, we have
\( \displaystyle{
\begin{array}{rcl}
h'(x) & = & (x^2 + 5x + 7)\frac{d}{dx}[x^3 + 2x  4] + \\
& & ~~~(x^3 + 2x  4) \frac{d}{dx}[x^2 + 5x + 7] \\
& = & (x^2 + 5x + 7)(3x^2 + 2) + \\
& & ~~~(x^3 + 2x  4)(2x + 5) \\
& = & (x^2)(3x^2 + 2) + 5x(3x^2 + 2) + \\
& & ~~~7(3x^2 + 2) + (x^3)(2x+5) + \\
& & ~~~ 2x(2x+5)  4(2x+5) \\
& = & 3x^4 + 2x^2 + 15x^3 + 10x + \\
& & ~~~21x^2 + 14 + 2x^4 + 5x^3 + \\
& & ~~~ 4x^2 + 10x  8x  20 \\
& = & 5x^4 + 20x^3 + 27x^2 + 12x  6
\end{array}
}\)
The product rule was used in the first step. The last three steps are just algebra. Keep in mind that we would normally check for a common factor after all the derivatives are done but before multiplying out. In this case, there is no common factor, so multiplying out is our only option. Leaving it in the form in the second line is not usually considered proper form (check with your instructor to see what they expect).
Practice A04 Final Answer 
\(h'(x)=5x^4+20x^3+27x^2+12x6\) 
Practice A05 
\(g(x)=(x^37x^2+4)(3x^2+14)\) 


\(g'(x)=x(15x^384x^2+42x172)\) 
\(\displaystyle{
\begin{array}{rcl}
\frac{dg}{dx}&=&(x^37x^2+4)\cdot\frac{d}{dx}[3x^2+14]+\\
& & ~~~ (3x^2+14)\cdot\frac{d}{dx}[x^37x^2+4]\\
&=&(x^37x^2+4)(6x)+(3x^2+14)(3x^214x)\\
&=&x[6x^342x^2+24+(3x^2+14)(3x14)]\\
&=&x(15x^384x^2+42x172)
\end{array}}\)
In the first line above we used the product rule. The following lines are just algebra. Note that we factored out an x before multiplying out between the 2nd and 3rd lines. This is a great time to factor. If we had multiplied out first and there were factors other than a single x term, they would have been difficult to see in the completely multiplied out form.
Practice A05 Final Answer 
\(g'(x)=x(15x^384x^2+42x172)\) 
Practice A06 
\(g(x)=(3x^4+2x1)(x^52x^2)\) 

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


\(\displaystyle{\frac{d}{dx}\left[x^3e^x\right]=x^2e^x(x+3)}\) 
We need to use the product rule here. So we have
\(\displaystyle{\begin{array}{rcl}
\frac{d}{dx}\left[x^3e^x\right]&=&x^3\cdot\frac{d[e^x]}{dx}+e^x\cdot\frac{d[x^3]}{dx}\\
&=&x^3\cdot e^x+e^x\cdot3x^2\\
&=&x^2e^x(x+3)\end{array}}\)
The product rule was used in the first line. Notice that our final answer is completely factored.
Practice A07 Final Answer 
\(\displaystyle{\frac{d}{dx}\left[x^3e^x\right]=x^2e^x(x+3)}\) 
Practice A08 
\(\displaystyle{y=7e^{2x}}\) 


\(\displaystyle{\frac{d}{dx}\left[7e^{2x}\right]=14e^{2x}}\) 
\( \displaystyle{
\begin{array}{rcl}
\frac{d}{dx} \left[ 7e^{2x} \right] & = & 7 \frac{d}{dx}[ e^x \cdot e^x ] \\
& = & 7e^x \frac{d[e^x]}{dx} + 7 \frac{d[e^x]}{dx}e^x \\
& = & 7e^x (e^x) + 7e^x (e^x) \\
& = & 14e^{2x}
\end{array}}\)
Note: The derivative is more easily solved 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 
\(\displaystyle{e^x\sqrt{x}}\) 


\( \displaystyle{\frac{d}{dx}\left[ e^x \sqrt{x} \right] = \left[ \frac{e^x}{2x^{1/2}} \right] (1 + 2x)} \) 
\( \displaystyle{
\begin{array}{rcl}
\frac{d}{dx}\left[ e^x \sqrt{x} \right] & = & \frac{d}{dx}\left[ e^x \left( x^{1/2} \right) \right] \\
& = & e^x \cdot \frac{d[x^{1/2}]}{dx} + \left( x^{1/2} \right) \cdot \frac{d[e^x]}{dx} \\
& = & e^x \cdot (1/2)x^{1/2} + x^{1/2} \cdot e^x \\
& = & \left[ \frac{e^x}{2x^{1/2}} \right] (1 + 2x)
\end{array}}\)
Notes
Rewriting \(\sqrt{x}\) as \(x^{1/2}\) allowed us to use the power rule.
In the first step, we used the product rule. After that, 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 for exponential terms. Your teacher 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 
\(\displaystyle{f(x)=\cos(x)\sin(x)}\) 


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


\(\displaystyle{\frac{d}{dx}\left[(x^2+3x5)\sqrt[3]{x}\right]=\frac{7x^2+12x5}{3x^{2/3}}}\) 
\( \displaystyle{ \begin{array}{rcl}
\frac{dg}{dx} & = & \frac{d}{dx}\left[ (x^2+3x5)(x^{1/3}) \right] \\
& = & (x^2 + 3x  5) \cdot \frac{d[x^{1/3}]}{dx} + \left( x^{1/3} \right) \cdot \frac{d[x^2+3x5]}{dx} \\
& = & (x^2 + 3x  5) \cdot (1/3)x^{2/3} + x^{1/3} \cdot (2x+3) \\
& = & \frac{ x^2 + 3x  5}{ 3x^{2/3}} + \left[ x^{1/3} \cdot (2x+3) \right] \left[\frac{3x^{2/3}}{3x^{2/3}} \right] \\
& = & \frac{ x^2 + 3x  5 + (3x)(2x+3) }{3x^{2/3}} \\
& = & \frac{7x^2 + 12x  5}{3x^{2/3}}
\end{array}}\)
Notes
Rewriting \(\sqrt[3]{x}\) as \( x^{1/3} \) allows us to more easily use the power rule.
In the first step, we used the product rule. After that, 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 teacher 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 teacher 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 
\(\displaystyle{\frac{3}{x}\cot(x)}\) 


\(\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] = \frac{3}{x^2}\left[ x \csc^2(x) + \cot(x) \right] }\) 
\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] & = & \frac{3}{x}\frac{d}{dx}[\cot(x)] + \cot(x) \frac{d}{dx}[3/x] \\
& = & \frac{3}{x} (\csc^2(x)) + \cot(x) \frac{d}{dx}[3x^{1}] \\
& = & \frac{3}{x} \csc^2(x) + \cot(x) [3(1)x^{2}] \\
& = & \frac{3}{x} \csc^2(x)  \frac{3}{x^2} \cot(x) \\
& = & \frac{3}{x^2}[ x \csc^2(x) + \cot(x) ]
\end{array}}\)
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 
\(\displaystyle{f(x)=4x^2e^x\sin(x)\sec(x)}\) 
