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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #11 : First And Second Derivatives Of Functions

Find the derivative of the following function:

$f(x)=\sec(3x)+e^{2x^2+1}$

Possible Answers:

$3\sec(3x)\tan(3x)+4xe^{2x^2+1}$

$\sec(3x)\tan(3x)+4xe^{2x^2+1}$

$3\sec(3x)\tan(3x)+e^{2x^2+1}$

$3\sec(x)\tan(x)+4xe^{2x^2+1}$

$3\sec(3x)\tan(3x)-e^{2x^2+1}$

Correct answer:

$3\sec(3x)\tan(3x)+4xe^{2x^2+1}$

Explanation:

The derivative of the function is equal to

$f'(x)=3\sec(3x)\tan(3x)+4xe^{2x^2+1}$

and was found using the following rules:

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

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Example Question #1 : Applications Of Derivatives

Let

$f(x)=e^{\pi x}$

Find the first and second derivative of the function.

Possible Answers:

$f'(x)=e^{\pi x }$

$f''(x)=e^{\pi x}$

$f'(x)=\pi e^{\pi x }$

$f''(x)=\pi^2 e^{\pi x}$

$f'(x)=e^{\pi }$

$f''(x)= e^{x}$

$f'(x)=\pi$

$f''(x)=\0$

Correct answer:

$f'(x)=\pi e^{\pi x }$

$f''(x)=\pi^2 e^{\pi x}$

Explanation:

In order to solve for the first and second derivative, we must use the chain rule.

The chain rule states that if

y=f(u)

and

u=g(x)

then the derivative is

$\frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}$

In order to find the first derviative of the function

$f(x)=e^{\pi x}$

we set

y=e^u

and

$u=\pi x$

Because the derivative of the exponential function is the exponential function itself, we get

$\frac{dy}{du}=\frac{d}{du}[e^u]$

$\frac{dy}{du}=e^u$

And differentiating we use the power rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

As such

$\frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]$

$\frac{du}{dx}=\pi x^{1-1}=\pi$

And so

$\frac{dy}{dx}=e^{\pi x}*\pi=\pi e^{\pi x}$

$f'(x)=\pi e^{\pi x}$

To solve for the second derivative we set

$y=\pi e^{u}$

and

$u=\pi x$

Because the derivative of the exponential function is the exponential function itself, we get

$\frac{dy}{du}=\frac{d}{du}[\pi e^u]$

$\frac{dy}{du}=\pi e^u$

And differentiating we use the power rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

As such

$\frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]$

$\frac{du}{dx}=\pi x^{1-1}=\pi$

And so the second derivative becomes

$\frac{dy}{dx}=\pi e^{\pi x}*\pi=\pi^2 e^{\pi x}$

$f''(x)=\pi^2e^{\pi x}$

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Example Question #212 : Derivative Review

Find the derivative of the following function:

$f(x)=\cos(e^{3x})+\tan(2x)$

Possible Answers:

$3e^{3x}\sin(e^{3x})-\sec^2(2x)$

$-\sin(e^{3x})+\sec^2(2x)$

$e^{3x}\sin(e^{3x})+2\sec^2(2x)$

$-3e^{3x}\sin(e^{3x})+2\sec^2(2x)$

Correct answer:

$-3e^{3x}\sin(e^{3x})+2\sec^2(2x)$

Explanation:

The derivative of the function is equal to

$f'(x)=-3e^{3x}\sin(e^{3x})+2\sec^2(2x)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$

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Example Question #213 : Derivative Review

Find the derivative of the function:

$f(x)=\frac{1}{3x\sqrt{x}}+\sec(x)$

Possible Answers:

$\frac{-6}{\sqrt{x}}+\ln\left |\sec(x)+\tan(x) \right |+C$

$-\frac{6}{\sqrt{x}}+\sec(x)\tan(x)+C$

$\frac{-6}{\sqrt{x}}+\ln\left |\sec(x)\tan(x) \right |+C$

$\frac{-3}{2\sqrt{x}}+\sec(x)\tan(x)$

$\frac{-6}{\sqrt{x}}-\ln\left |\sec(x)+\tan(x) \right |+C$

Correct answer:

$\frac{-6}{\sqrt{x}}+\ln\left |\sec(x)+\tan(x) \right |+C$

Explanation:

The integral of the function is equal to

$\frac{-6}{\sqrt{x}}+\ln\left |\sec(x)+\tan(x) \right |+C$

and was found using the following rules:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$ , $\int \sec(x)dx=\ln\left | \sec(x)+\tan(x)\right |+C$

Note that it is easier to integrate the first term once it is rewritten as $3x^{-\frac{3}{2}}$ .

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Example Question #2 : Applications Of Derivatives

Find the velocity function of the particle if its position is given by the following function:

$s(t)=e^t\tan(t)+t^5+3$

Possible Answers:

$te^t\tan(t)+e^t\sec^2(t)+5t^4$

$e^t\tan(t)+e^t\sec^2(t)+t^5$

$e^t\tan(t)-e^t\sec^2(t)+5t^4$

$e^t\tan(t)+e^t\sec^2(t)+5t^4$

Correct answer:

$e^t\tan(t)+e^t\sec^2(t)+5t^4$

Explanation:

The velocity function is given by the first derivative of the position function:

$v(t)=s'(t)=e^t\tan(t)+e^t\sec^2(t)+5t^4$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(t)=\sec^2(t)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$

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Example Question #12 : First And Second Derivatives Of Functions

Find the second derivative of the following function:

$f(x)=x^3\cos(x)+e^{2x}$

Possible Answers:

$\cos(x)(6x-x^3)-6x^2\sin(x)+2e^{2x}$

$\cos(x)(6x-x^3)+6x^2\sin(x)+4e^{2x}$

$\cos(x)(6x-x^3)-6x^2\sin(x)+4e^{2x}$

$\cos(x)(6x+x^3)-6x^2\sin(x)+4e^{2x}$

$\cos(x)(6x-x^3)-3x^2\sin(x)+4e^{2x}\:$

Correct answer:

$\cos(x)(6x-x^3)-6x^2\sin(x)+4e^{2x}$

Explanation:

The first derivative of the function is equal to

$f'(x)=3x^2\cos(x)-x^3\sin(x)+2e^{2x}$

The second derivative - the derivative of the function above - of the original function is equal to

$f''(x)=\cos(x)(6x-x^3)-6x^2\sin(x)+4e^{2x}$

Both derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}$

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Example Question #1341 : Calculus Ii

Find the derivative of the following function:

$f(x)=\sqrt[3]{\cos^2(x)+e^{4x}}$

Possible Answers:

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

$\frac{-2\cos(x)\sin(x)+4e^{4x}}{3(\cos^2(x)+e^{4x})^{\frac{1}{3}}}$

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

$\frac{2\cos(x)\sin(x)+4e^{4x}}{3(\cos^2(x)+e^{4x})^{\frac{2}{3}}}$

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

Correct answer:

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

Explanation:

The derivative was found using the following rules:

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

$f(x)=\sqrt[3]{\cos^2(x)+e^{4x}}$

$f'(x)=\frac{1}{3}(\cos^2(x)+e^{4x})^{-\frac{2}{3}}(-2\cos(x)\sin(x)+4e^{4x})$

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

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Example Question #1342 : Calculus Ii

Find the derivative of the function:

$f(x)=\frac{e^{2x}+\cos(5x^2)}{e^x}$

Possible Answers:

$\frac{e^{3x}-10x\sin(5x^2)+\cos(5x^2)}{e^x}$

$\frac{e^{2x}-10x\sin(5x^2)+\cos(5x^2)}{e^x}$

$\frac{e^{2x}-10xe^x\sin(5x^2)+\cos(5x^2)}{e^x}$

$\frac{e^{2x}+10x\sin(5x^2)+\cos(5x^2)}{e^x}$

Correct answer:

$\frac{e^{2x}-10x\sin(5x^2)+\cos(5x^2)}{e^x}$

Explanation:

The derivative of the function is equal to

$f(x)=\frac{e^{2x}+\cos(5x^2)}{e^x}$

$f'(x)=\frac{(2e^{2x}-10x\sin(5x^2))e^x-e^x(e^{2x}+\cos(5x^2))}{e^{2x}}$

$f'(x)=\frac{e^{2x}-10x\sin(5x^2)+\cos(5x^2)}{e^x}$

and was found using the following rules:

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

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Example Question #212 : Derivatives

Find the derivative of the following function:

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

$f'(x)=\frac{1}{2}(e^{\cos(x)})^{-\frac{1}{2}}(-\sin(x)e^{\cos(x)})$

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

The derivative of the function was found using the following rules:

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

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Example Question #1344 : Calculus Ii

Solve for the derivative: y=2cot(2x)

Possible Answers:

-4csc({2x})

$-4sec^2({2x})$

-4sec({2x})tan({2x})

$-2sec^2({2x})$

$-4csc^2({2x})$

Correct answer:

$-4csc^2({2x})$

Explanation:

Write the derivative of cotangent .

cot(x) = -csc^2(x)

The problem requires chain rule since the inner function is . The chain rule is the derivative of the inner function, and the derivative of is .

Take the derivative to obtain $\frac{dy}{dx}$ . Multiply the value obtained by the use of chain rule.

$\frac{dy}{dx} = 2 (-csc^2({2x})) \cdot 2 = -4csc^2({2x})$

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #11 : First And Second Derivatives Of Functions

Example Question #1 : Applications Of Derivatives

Example Question #212 : Derivative Review

Example Question #213 : Derivative Review

Example Question #2 : Applications Of Derivatives

Example Question #12 : First And Second Derivatives Of Functions

Example Question #1341 : Calculus Ii

Example Question #1342 : Calculus Ii

Example Question #212 : Derivatives

Example Question #1344 : Calculus Ii

All Calculus 2 Resources

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