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Calculus 2 : Derivative Review

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Example Questions

Example Question #221 : Derivative Review

Use the chain rule to find the derivative of the function $f(x) = e^{-6x^2}$

Possible Answers:

$-6x^2e^{-6x^2}$

-6x^2e

$-12xe^{-6x^2}$

$e^{-6x^2}$

None of these answers

Correct answer:

$-12xe^{-6x^2}$

Explanation:

The formula for the chain rule works as follows;

$[f(g(x))]' = f'(g(x))\times g'(x)$

Setting f(x) = e^x, g(x)=-6x^2 , we have

$f(g(x)) = e^{-6x^2}$

Hence

$[f(g(x))]' = f'(g(x))\times g'(x) = e^{-6x^2} \times -12x = -12xe^{-6x^2}$ .

(Keep in mind that the derivative of e^x is e^x )

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Example Question #22 : First And Second Derivatives Of Functions

Find the second derivative of: y= sec(2x)

Possible Answers:

8sec(2x)tan^2(2x)+ 8sec^3(2x)

2sec^3(2x)+2sec(2x)tan^2(2x)

2sec^2(2x)+2sec(2x)tan(2x)

4sec^3(2x)+4sec(2x)tan^2(2x)

8sec(4x)tan(4x) +4sec^3(2x)

Correct answer:

4sec^3(2x)+4sec(2x)tan^2(2x)

Explanation:

The derivative of secant is:

$\frac{d}{dx}sec(x) = sec(x)tan(x)$

Since the inner function of secant is not , we will need to use chain rule, which is to multiply the derivative of the inner function .

Solve for the first derivative.

$\frac{d}{dx}sec(2x) = sec(2x)tan(2x) \cdot 2 = 2sec(2x)tan(2x)$

To take the second derivative, we will need to use the product rule and chain rule.

The product rule is: f(x) g'(x) + g(x)f'(x)

To avoid confusion, let f(x) =sec(2x) and g(x)= tan(2x) . Their derivatives are then:

f'(x) = 2sec(2x)tan(2x)

g'(x) = 2sec^2(2x)

The derivative of 2sec(2x)tan(2x) in terms of f(x) and g(x) by product rule is:

$\frac{d^2}{dx^2}:\:2[f(x) g'(x) + g(x)f'(x)]$

Resubstitute the functions and derivative functions into the equation.

$2[sec(2x)\cdot 2sec^2(2x)+ tan(2x)\cdot 2sec(2x)tan(2x)]$

Simplify. The answer is:

4sec^3(2x)+4sec(2x)tan^2(2x)

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Example Question #22 : First And Second Derivatives Of Functions

Find the derivative of: $y=sec(\frac{1}{x})$

Possible Answers:

$\frac{sec(\frac{1}{x})tan(\frac{1}{x})}{x^2}$

$-\frac{sec(\frac{1}{x})tan(\frac{1}{x})}{x^2}$

$-ln(x)sec(\frac{1}{x})tan(\frac{1}{x})$

$ln(x)sec(\frac{1}{x})tan(\frac{1}{x})$

$sec(\frac{1}{x})tan(\frac{1}{x})$

Correct answer:

$-\frac{sec(\frac{1}{x})tan(\frac{1}{x})}{x^2}$

Explanation:

Write the derivative of secant.

$\frac{d}{dx} sec(x) = sec(x)tan(x)$

Since the inner function is $\frac{1}{x}$ , we will need to apply the chain rule to solve this derivative.

Take the derivative of $\frac{1}{x}$ . Do not mix this with the integration, which is ln(x) .

$\frac{d}{dx}\:\frac{1}{x} = \frac{d}{dx}\:x^{-1} = -1\cdot x^{-2} = -\frac{1}{x^2}$

The derivative of sec(x) is then:

$= sec(\frac{1}{x})tan(\frac{1}{x}) \cdot -\frac{1}{x^2}$

The answer is: $-\frac{sec(\frac{1}{x})tan(\frac{1}{x})}{x^2}$

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Example Question #21 : First And Second Derivatives Of Functions

What is the second derivative of $\sin(x^2)$ ?

Possible Answers:

$-\sin(x^2)$

$2\cos(x^2)-4x^2\cdot \sin(x^2)$

$2x\cdot \cos(x^2)$

$2x\cdot \cos(x^2) - x^4\cdot \sin(x^2)$

Correct answer:

$2\cos(x^2)-4x^2\cdot \sin(x^2)$

Explanation:

We use the chain rule to get the first derivative:

$(f \circ g)' = (f'\circ g)\cdot g'$

This gives us $2x\cdot \cos(x^2)$ . Now we must take the derivative again, which means we have to use the product rule *and* the chain rule. As a reminder, the product rule says that

$(f\cdot g)' = f'\cdot g + f\cdot g'$

Which means the second derivative of the original function is

$2\cos(x^2)+2x\cdot 2x (-\sin(x^2))$ . With some simplifying we can see that this is equal to

$2\cos(x^2)-4x^2\sin(x^2)$

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Example Question #23 : First And Second Derivatives Of Functions

Evaluate the first and second derivative of the function

$f(x)=\pi x$

when x=4

Possible Answers:

$f'(4)=4\pi$

f''(4)=4

$f'(4)=4x\pi$

$f''(4)=4\pi$

$f'(4)=\pi$

f''(4)=0

f'(4)=4x

f''(4)=x

Correct answer:

$f'(4)=\pi$

f''(4)=0

Explanation:

We must find the first and second derivatives.

To do so we use the power rule which states

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

As such

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]=1*\pi x^{1-1}=\pi x^0 = \pi$

To find the second derivative we apply the power rule again

$f''(x)=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x^0]=0*\pi x^{0-1}=0$

We find that

$f'(x)=\pi$ and f''(x)=0

We then evaluate each for x=4 and get

$f'(4)=\pi$

f''(4)=0

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Example Question #1 : Applications Of Derivatives

Find the first and second derivatives of the function

f(x)=sec(x)

Possible Answers:

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

f'(x)=sec^2(x)

f''(x)=sec^3(x)

f'(x)=csc(x)

f''(x)=-sec(x)

f'(x)=cot(x)

f''(x)=csc^2(x)

Correct answer:

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

Explanation:

We must find the first and second derivatives.

We use the properties that

The derivative of is
The derivative of is

As such

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[sec(x)]=sec(x)tan(x)$

To find the second derivative we differentiate again and use the product rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}[uv]=uv'+vu'$

Setting

u=sec(x) and v=tan(x)

we find that

$f''(x)=\frac{\mathrm{d} }{\mathrm{d} x}[sec(x)tan(x)]=sec(x)*sec^2(x)+tan(x)*sec(x)tan(x)$

=sec^3(x)+sec(x)tan^2(x)

As such

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

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Example Question #21 : First And Second Derivatives Of Functions

Find the derivative of the following function:

$f(x)=\frac{x^3}{\sin(x)}+\ln(x^2)$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

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

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Example Question #21 : First And Second Derivatives Of Functions

Find the derivative of the function:

$f(x)=\cos(e^{2x})+x^5+3x$

Possible Answers:

$-2e^{2x}\sin(e^{2x})+5x^4+3x$

$-2e^{2x}\sin(e^{2x})+5x^4+3$

$2e^{2x}\sin(e^{2x})+5x^4+3$

$2e^{2x}\cos(e^{2x})+5x^4+3$

Correct answer:

$-2e^{2x}\sin(e^{2x})+5x^4+3$

Explanation:

The derivative of the function is equal to

$f'(x)=-2e^{2x}\sin(e^{2x})+5x^4+3$

and was found using the following rules:

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

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Example Question #223 : Derivative Review

Find the second derivative of the function:

$\ln(5x)+\cos(x^2)$

Possible Answers:

$-\frac{5}{x^2}-2\sin(x^2)+4x^2\cos(x^2)$

$\frac{5}{x^2}-2\sin(x^2)-4x^2\cos(x^2)$

$-\frac{5}{x^2}-\sin(x^2)-4x^2\cos(x^2)$

$-\frac{5}{x^2}-2\sin(x^2)-4x^2\cos(x^2)$

Correct answer:

$-\frac{5}{x^2}-2\sin(x^2)-4x^2\cos(x^2)$

Explanation:

The first derivative of the function is equal to

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

and was found using the following rules:

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

The second derivative of the function is equal to

$f''(x)=-\frac{5}{x^2}-2\sin(x^2)-4x^2\cos(x^2)$

and was found using the same rules as above, along with

$\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$

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Example Question #131 : Derivatives

Find the derivative of the function

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

Possible Answers:

$\frac{1-\ln(x)^2}{[\ln(x)]^4}$

$\frac{(x^2)(1-\ln(x)^2)}{[\ln(x)]^4}$

$\frac{\ln(x)-1}{[\ln(x)]^2}$

$\frac{1-\ln(x)}{[\ln(x)]^2}$

None of the other answers.

Correct answer:

$\frac{\ln(x)-1}{[\ln(x)]^2}$

Explanation:

We find the answer using the quotient rule

$(\frac{f(x)}{g(x)})' = \frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$

and the product rule

(f(x)g(x))' = f(x)g'(x)+g(x)f'(x)

and then simplifying.

$y' = \frac{x\ln(x)\times 2x-x^2\times(x\frac{1}{x}+\ln(x)))}{(x\ln(x))^2}$

$= \frac{2x^2\ln(x)-x^2-x^2\ln(x)}{x^2\ln(x)^2}$

$=\frac{2\ln(x)-1-\ln(x)}{\ln(x)^2}$

$=\frac{\ln(x)-1}{\ln(x)^2}$ or $\frac{\ln(x)-1}{[\ln(x)]^2}$ . The extra brackets in the denominator are optional.

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #221 : Derivative Review

Example Question #22 : First And Second Derivatives Of Functions

Example Question #22 : First And Second Derivatives Of Functions

Example Question #21 : First And Second Derivatives Of Functions

Example Question #23 : First And Second Derivatives Of Functions

Example Question #1 : Applications Of Derivatives

Example Question #21 : First And Second Derivatives Of Functions

Example Question #21 : First And Second Derivatives Of Functions

Example Question #223 : Derivative Review

Example Question #131 : Derivatives

All Calculus 2 Resources

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