You CAN Ace Calculus
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. |
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. |
external links you may find helpful |
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Single Variable Calculus |
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Multi-Variable Calculus |
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Acceleration Vector |
Arc Length (Vector Functions) |
Arc Length Function |
Arc Length Parameter |
Conservative Vector Fields |
Cross Product |
Curl |
Curvature |
Cylindrical Coordinates |
Lagrange Multipliers |
Line Integrals |
Partial Derivatives |
Partial Integrals |
Path Integrals |
Potential Functions |
Principal Unit Normal Vector |
Differential Equations |
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Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
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Product Rule |
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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.
video by PatrickJMT
Okay, practice problem time. |
quotient rule → |
Instructions - - Unless otherwise instructed, calculate the derivatives of the following functions using the product rule, giving your final answers in simplified, factored form.
Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems |
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Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. Here is a list converting the old numbering system to the new. |
Derivative Product Rule - Practice Problems Conversion |
[A01-938] - [A02-1928] - [A03-939] - [A04-925] - [A05-927] - [A06-1311] - [A07-928] - [A08-929] - [A09-930] |
[A10-931] - [A11-932] - [B01-926] - [B02-935] - [B03-936] - [B04-940] - [B05-937] - [B06-933] - [C01-934] |
Please update your notes to this new numbering system. The display of this conversion information is temporary. |
Basic Problems |
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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 |
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Calculate the derivative of \( f(x) = x(25-x) \) using the product rule.
Solution |
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video by Krista King Math
close solution |
\( f(x) = (x-2)(x+3) \)
Problem Statement |
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Calculate the derivative of \( f(x) = (x-2)(x+3) \) using the product rule.
Final Answer |
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\(f'(x)=2x+1\) |
Problem Statement |
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Calculate the derivative of \( f(x) = (x-2)(x+3) \) using the product rule.
Solution |
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video by PatrickJMT
Final Answer |
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\(f'(x)=2x+1\) |
close solution |
\( f(x) = (2x+3)(3x-2) \)
Problem Statement |
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Calculate the derivative of \( f(x) = (2x+3)(3x-2) \) using the product rule.
Solution |
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video by Krista King Math
close solution |
\(h(x)=(x^2+5x+7)(x^3+2x-4)\)
Problem Statement |
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Calculate the derivative of \( h(x) = (x^2+5x+7)(x^3+2x-4) \) using the product rule.
Final Answer |
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\(h'(x)=5x^4+20x^3+27x^2+12x-6\) |
Problem Statement |
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Calculate the derivative of \( h(x) = (x^2+5x+7)(x^3+2x-4) \) using the product rule.
Solution |
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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. |
\( h'(x) = 5x^4 + 20x^3 + 27x^2 + 12x - 6 \) |
Final Answer |
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\(h'(x)=5x^4+20x^3+27x^2+12x-6\) |
close solution |
\( g(x) = (x^3-7x^2+4)\) \((3x^2+14) \)
Problem Statement |
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Calculate the derivative of \( g(x) = (x^3-7x^2+4)\) \((3x^2+14) \) using the product rule.
Final Answer |
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\( g'(x) = x(15x^3-84x^2+42x-172) \) |
Problem Statement |
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Calculate the derivative of \( g(x) = (x^3-7x^2+4)\) \((3x^2+14) \) using the product rule.
Solution |
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First, we apply the product rule. |
\( \displaystyle{ \frac{dg}{dx} = }\) \(\displaystyle{ (x^3-7x^2+4)\cdot }\) \(\displaystyle{ \frac{d}{dx}[3x^2+14]+ }\) \(\displaystyle{ (3x^2+14)\cdot }\) \(\displaystyle{ \frac{d}{dx}[x^3-7x^2+4] }\) |
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.
Final Answer |
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\( g'(x) = x(15x^3-84x^2+42x-172) \) |
close solution |
\( g(x) = (3x^4+2x-1)\) \((x^5-2x^2) \)
Problem Statement |
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Calculate the derivative of \( g(x) = (3x^4+2x-1)\) \((x^5-2x^2) \) using the product rule.
Solution |
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video by PatrickJMT
close solution |
These problems require you to know how to take the derivative of exponential functions.
\(\displaystyle{f(x)=x^3e^x}\)
Problem Statement |
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Calculate the derivative of \(\displaystyle{f(x)=x^3e^x}\) using the product rule.
Final Answer |
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\(\displaystyle{\frac{d}{dx}\left[x^3e^x\right]=x^2e^x(x+3)}\) |
Problem Statement |
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Calculate the derivative of \(\displaystyle{f(x)=x^3e^x}\) using the product rule.
Solution |
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\(\displaystyle{ \frac{d}{dx}\left[x^3e^x\right] }\) |
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. |
\( x^2e^x(x+3) \) |
Final Answer |
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\(\displaystyle{\frac{d}{dx}\left[x^3e^x\right]=x^2e^x(x+3)}\) |
close solution |
\(\displaystyle{y=7e^{2x}}\)
Problem Statement |
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Calculate the derivative of \(\displaystyle{y=7e^{2x}}\) using the product rule.
Final Answer |
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\(\displaystyle{\frac{d}{dx}\left[7e^{2x}\right]=14e^{2x}}\) |
Problem Statement |
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Calculate the derivative of \(\displaystyle{y=7e^{2x}}\) using the product rule.
Solution |
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\(\displaystyle{ \frac{d}{dx} \left[ 7e^{2x} \right] }\) |
\(\displaystyle{ 7 \frac{d}{dx}[ e^x \cdot e^x ] }\) |
\(\displaystyle{ 7e^x \frac{d[e^x]}{dx} + 7e^x \frac{d[e^x]}{dx} }\) |
\( 7e^x (e^x) + 7e^x (e^x) \) |
\( 14e^{2x} \) |
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.
Final Answer |
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\(\displaystyle{\frac{d}{dx}\left[7e^{2x}\right]=14e^{2x}}\) |
close solution |
\(\displaystyle{e^x\sqrt{x}}\)
Problem Statement |
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Calculate the derivative of \(\displaystyle{e^x\sqrt{x}}\) using the product rule.
Final Answer |
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\( \displaystyle{ \frac{e^x (1 + 2x)}{2x^{1/2}} } \) |
Problem Statement |
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Calculate the derivative of \(\displaystyle{e^x\sqrt{x}}\) using the product rule.
Solution |
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\(\displaystyle{ \frac{d}{dx}\left[ e^x \sqrt{x} \right] }\) |
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.
Final Answer |
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\( \displaystyle{ \frac{e^x (1 + 2x)}{2x^{1/2}} } \) |
close solution |
These problems require you to know how to take the derivative of trig functions.
\(\displaystyle{x^4\tan(x)}\)
Problem Statement |
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Calculate the derivative of \(\displaystyle{x^4\tan(x)}\) using the product rule.
Final Answer |
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\(\displaystyle{ x^3[x \sec^2(x)+4 \tan(x)] }\) |
Problem Statement |
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Calculate the derivative of \(\displaystyle{x^4\tan(x)}\) using the product rule.
Solution |
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\(\displaystyle{\frac{d}{dx}[ x^4 \tan(x) ] = x^4 \sec^2(x) + 4x^3 \tan(x) = }\) \(\displaystyle{x^3[x \sec^2(x)+4 \tan(x)]}\)
Note: Not factoring out the \(x^3\) term, may cost you points. Check with your instructor to see what they expect.
Final Answer |
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\(\displaystyle{ x^3[x \sec^2(x)+4 \tan(x)] }\) |
close solution |
\(\displaystyle{f(x)=\cos(x)\sin(x)}\)
Problem Statement |
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Calculate the derivative of \(\displaystyle{f(x)=\cos(x)\sin(x)}\) using the product rule.
Final Answer |
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\( f'(x) = \cos^2(x) - \sin^2(x) \) |
Problem Statement |
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Calculate the derivative of \(\displaystyle{f(x)=\cos(x)\sin(x)}\) using the product rule.
Solution |
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\(\displaystyle{ f'(x) = \frac{d}{dx}\left[ \cos(x) \sin(x) \right] }\) |
\(\displaystyle{ f'(x) = \cos(x) \frac{d}{dx}[ \sin(x)] + \sin(x) \frac{d}{dx}[ \cos(x)] }\) |
\( f'(x) = \cos(x)[ \cos(x)] + \sin(x)[-\sin(x)] \) |
\( f'(x) = \cos^2(x) - \sin^2(x) \) |
Final Answer |
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\( f'(x) = \cos^2(x) - \sin^2(x) \) |
close solution |
Intermediate Problems |
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\(\displaystyle{g(x)=(x^2+3x-5)\sqrt[3]{x}}\)
Problem Statement |
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Calculate the derivative of \(\displaystyle{g(x)=(x^2+3x-5)\sqrt[3]{x}}\) using the product rule.
Final Answer |
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\(\displaystyle{\frac{d}{dx}\left[(x^2+3x-5)\sqrt[3]{x}\right]=\frac{7x^2+12x-5}{3x^{2/3}}}\) |
Problem Statement |
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Calculate the derivative of \(\displaystyle{g(x)=(x^2+3x-5)\sqrt[3]{x}}\) using the product rule.
Solution |
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Rewrite \(\sqrt[3]{x}\) 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 \(\sqrt[3]{x^2}\). 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.
Final Answer |
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\(\displaystyle{\frac{d}{dx}\left[(x^2+3x-5)\sqrt[3]{x}\right]=\frac{7x^2+12x-5}{3x^{2/3}}}\) |
close solution |
\(\displaystyle{f(x)=\left[\frac{1}{x^2}-\frac{3}{x^4}\right]\cdot}\) \((x+5x^3)\)
Problem Statement |
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Calculate the derivative of \(\displaystyle{f(x)=\left[\frac{1}{x^2}-\frac{3}{x^4}\right]\cdot}\) \((x+5x^3)\) using the product rule.
Solution |
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video by Krista King Math
close solution |
\( H(u)=(u-\sqrt{u}) \) \( (u+\sqrt{u}) \)
Problem Statement |
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Calculate the derivative of \( H(u)=(u-\sqrt{u}) \) \( (u+\sqrt{u}) \) using the product rule.
Solution |
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video by Krista King Math
close solution |
\(f(x)=x(2x-1)\) \((2x+1)\)
Problem Statement |
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Calculate the derivative of \(f(x)=x(2x-1)\) \((2x+1)\) using the product rule.
Solution |
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video by Krista King Math
close solution |
\(f(x)=(2x+1)\) \((4-x^2)(1+x^2)\)
Problem Statement |
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Calculate the derivative of \(f(x)=(2x+1)\) \((4-x^2)(1+x^2)\) using the product rule.
Solution |
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video by Krista King Math
close solution |
Here is one with a trig derivative.
\(\displaystyle{\frac{3}{x}\cot(x)}\)
Problem Statement |
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Calculate the derivative of \(\displaystyle{\frac{3}{x}\cot(x)}\) using the product rule.
Final Answer |
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\(\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] = \frac{-3}{x^2}\left[ x \csc^2(x) + \cot(x) \right] }\) |
Problem Statement |
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Calculate the derivative of \(\displaystyle{\frac{3}{x}\cot(x)}\) using the product rule.
Solution |
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\(\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] }\) |
\(\displaystyle{ \frac{3}{x}\frac{d}{dx}[\cot(x)] + \cot(x) \frac{d}{dx}[3/x] }\) |
\(\displaystyle{ \frac{3}{x} (-\csc^2(x)) + \cot(x) \frac{d}{dx}[3x^{-1}] }\) |
\(\displaystyle{ \frac{-3}{x} \csc^2(x) + \cot(x) [3(-1)x^{-2}] }\) |
\(\displaystyle{ \frac{-3}{x} \csc^2(x) - \frac{3}{x^2} \cot(x) }\) |
\(\displaystyle{ \frac{-3}{x^2}[ x \csc^2(x) + \cot(x) ] }\) |
Final Answer |
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\(\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] = \frac{-3}{x^2}\left[ x \csc^2(x) + \cot(x) \right] }\) |
close solution |
Advanced Problems |
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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 |
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Calculate the derivative of \(f(x)=4x^2e^x\) \(\sin(x)\sec(x)\) using the product rule.
Solution |
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video by Krista King Math
close solution |