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Calculus 2 : Derivative Review

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Example Questions

Example Question #321 : Derivative Review

Find the derivative of f(x)=-3x^3-9x^2+4x-1 .

Possible Answers:

$f{}'(x)=-27x^2-18x$

$f{}'(x)=27x^2-18x+4$

$f{}'(x)=-27x^2+18x+4$

$f{}'(x)=-27x^2-18x+4x$

$f{}'(x)=-27x^2-18x+4$

Correct answer:

$f{}'(x)=-27x^2-18x+4$

Explanation:

To take the derivative, remember to multiply the exponent of an x term by the coefficient in front of the x term and then subtract one from the exponent:

$f{}'(x)=-27x^2-18x+4$ .

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Example Question #322 : Derivative Review

What is the acceleration function if x(t)=-16t^2+4t-1?

Possible Answers:

a(t)=-3

a(t)=-32

a(t)=-32t

a(t)=-32t^2

a(t)=-32t-1

Correct answer:

a(t)=-32

Explanation:

Recall that the acceleration function is the second derivative of the position function. First, find the velocity function since that is the first derivative. To take the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

v(t)=-32t+4

Now, take the derivative of the velocity function to get acceleration:

a(t)=-32

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Example Question #323 : Derivative Review

Find the first derivative of the function:

$f(x)=\tan(2xe^x)$

Possible Answers:

$\sec^2(2xe^x)[2e^x+2xe^x]$

$\sec^2(2xe^x)$

$\sec^2(2xe^x)[4e^x]$

$\sec^2(2xe^x)[2e^x+xe^x]$

Correct answer:

$\sec^2(2xe^x)[2e^x+2xe^x]$

Explanation:

The first derivative of the function is equal to

$f'(x)=\sec^2(2xe^x)[2e^x+2xe^x]$

and was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #324 : Derivative Review

Find the derivative of the following:

$f(x)=\frac{\cos(2x)}{e^{2x}}$

Possible Answers:

$\frac{-2(\sin(2x)+\cos(2x))}{e^{2x}}$

$-2(\sin(2x)-\cos(2x))$

$\frac{\sin(2x)-\cos(2x)}{e^{2x}}$

$\frac{-2(\sin(2x)-\cos(2x))}{e^{2x}}$

Correct answer:

$\frac{-2(\sin(2x)-\cos(2x))}{e^{2x}}$

Explanation:

The derivative of the function is equal to

$f'(x)=\frac{-2(\sin(2x)-\cos(2x))}{e^{2x}}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

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Example Question #325 : Derivative Review

Find the first derivative of the following function:

$f(x)=\frac{x^4}{\cos(x)}+\csc(3x)$

Possible Answers:

$\frac{4x^3\cos(x)-x^4\sin(x)}{\cos^2(x)}-3\csc(3x)\cot(3x)$

$\frac{4x^3\cos(x)-x^4\sin(x)}{\cos^2(x)}-\csc(3x)\cot(3x)$

$\frac{4x^3\cos(x)+x^4\sin(x)}{\cos^2(x)}-3\csc(3x)\cot(3x)$

$f'(x)=\frac{4x^3\cos(x)+x^4\sin(x)}{\cos^2(x)}+3\csc(3x)\cot(3x)$

Correct answer:

$\frac{4x^3\cos(x)+x^4\sin(x)}{\cos^2(x)}-3\csc(3x)\cot(3x)$

Explanation:

The first derivative of the function is equal to

$f'(x)=\frac{4x^3\cos(x)+x^4\sin(x)}{\cos^2(x)}-3\csc(3x)\cot(3x)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)$

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Example Question #322 : Derivative Review

Find the derivative of the function:

$f(x)=(x^2+3x)^{5/2}$

Possible Answers:

$\frac{5}{2}(2x+3)\sqrt{x^2+3x}$

$\frac{5}{2}(x^2+3x)^{\frac{3}{2}}(5x)$

$\frac{5}{2}(x^2+3x)(2x+3)$

$\frac{5}{2}(x^2+3x)^{\frac{3}{2}}(2x+3)$

Correct answer:

$\frac{5}{2}(x^2+3x)^{\frac{3}{2}}(2x+3)$

Explanation:

The derivative of the function is equal to

$f'(x)=\frac{5}{2}(x^2+3x)^{\frac{3}{2}}(2x+3)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Note that the exponent is the acting as the outer function in the chain rule.

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Example Question #323 : Derivative Review

Find the derivative of the function $f(x)= e^\sin(x)$

Possible Answers:

$\cos(x)e^\sin(x)$

$-e^\sin(x)$

$-\cos(x)e^\cos(x)$

$e^\sin(x)$

Correct answer:

$\cos(x)e^\sin(x)$

Explanation:

To find the derivative, you must first use the chain rule. The derivative of $e^\sin(x)$ is $e^\sin(x)$ , and you multiply that by the derivative of the exponent, which is $\cos(x)$ . Putting that all together, you get the final answer as $\cos(x)e^\sin(x)$ .

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Example Question #1453 : Calculus Ii

Find the second derivative of the function $f(x)= 3x^3+\sin(x)$ .

Possible Answers:

$9x^2+\cos(x)$

$18x+\sin(x)$

9x+cos(x)

$18x-\sin(x)$

Correct answer:

$18x-\sin(x)$

Explanation:

Taking the first derivative of the function, using the rule $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ and the fact that $\sin'(x)=\cos(x)$ , we get $f'(x)= 9x^2+\cos(x)$ . Using the same rule and the fact that $\cos'(x)=-\sin(x)$ , we get $f''(x)=18x-\sin(x)$

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Example Question #1454 : Calculus Ii

Find the derivative of the function $y(t)= \sin(t)+\frac{5}{t^2}$

Possible Answers:

$\cos(t)+8$

$\cos(t)-\frac{10}{t^3}$

$-\sin(t)+10t$

$\sin(t)+1$

Correct answer:

$\cos(t)-\frac{10}{t^3}$

Explanation:

To find the derivative of the function, we first take the derivative of $\sin(t)$ which is just $\cos(t)$ . Taking the derivative of the second term gets us $\frac{-10}{t^3}$ , using the rule $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ , where in this case, n=-2 . Using all of the information above and applying, $y'(t)= \cos(t)-\frac{10}{t^3}$

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Example Question #1455 : Calculus Ii

Find the second derivative of the function $f(x)=-\cos(x)$ . Evaluate at $x=\pi/2$ .

Possible Answers:

$-\frac{1}{2}$

Correct answer:

Explanation:

We first take the first derivative of the function which is $f'(x)=\sin(x)$ . We then take the derivative of this new function, which is $f''(x)=\cos(x)$ . Evaluating at $x=\pi/2$ , we get $\cos(\pi/2)=0$ .

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #321 : Derivative Review

Example Question #322 : Derivative Review

Example Question #323 : Derivative Review

Example Question #324 : Derivative Review

Example Question #325 : Derivative Review

Example Question #322 : Derivative Review

Example Question #323 : Derivative Review

Example Question #1453 : Calculus Ii

Example Question #1454 : Calculus Ii

Example Question #1455 : Calculus Ii

All Calculus 2 Resources

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