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Calculus 2 : Calculus II

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #43 : First And Second Derivatives Of Functions

Given the displacement function $f(x)=\ln(e^x+e^{2x})$ , find the velocity function.

Possible Answers:

$f'(x)=\frac{1}{e^x+e^{2x}}$

f'(x)=3

f'(x)=1

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

Correct answer:

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

Explanation:

To find the velocity of $f(x)=\ln(e^x+e^{2x})$ , we need to find the derivative. This can be done with the chain rule:

(g(h(x)))'=g'(h(x))h'(x)

with $g(x)=\ln x$ and $h(x)=e^x+e^{2x}$ , so we get

$g'(x)=\frac{1}{x}$

and

$h'(x)=e^x+e^{2x}$

so we get

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

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Example Question #241 : Derivatives

Given the displacement function

f(x)=ex+x^2

find the velocity function.

Possible Answers:

f'(x)=e^x+2x

f'(x)=e+2x

f'(x)=e+x^2

f'(x)=ex+2x

Correct answer:

f'(x)=e+2x

Explanation:

To find the velocity of f(x)=ex+x^2 , we need to find the derivative. This is simply:

f'(x)=e+2x

using the power rule:

$\frac{d}{dx}x^a=ax^{a-1}$

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Example Question #45 : First And Second Derivatives Of Functions

Given the velocity function

$v(x)=e^{5x}-e^{-5x}$

find the acceleration function a(x) .

Possible Answers:

$a(x)=e^{5x}-e^{-5x}$

$a(x)=5e^{5x}-5e^{-5x}$

$a(x)=5e^{5x}+5e^{-5x}$

Correct answer:

$a(x)=5e^{5x}+5e^{-5x}$

Explanation:

The acceleration function a(x) can be derived from the velocity function v(x) by taking the derivative: a(x)=v'(x) . So we get

$a(x)=v'(x)=\frac{d}{dx}(e^{5x}-e^{-5x})=5e^{5x}+5e^{-5x}$

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Example Question #46 : First And Second Derivatives Of Functions

Given the velocity function

$v(x)=e^{2x}-4e^{-6x}$

find the acceleration function a(x) .

Possible Answers:

$a(x)=e^{2x}+e^{-6x}$

$a(x)=2e^{2x}+24e^{-6x}$

$a(x)=2e^{2x}-24e^{-6x}$

Correct answer:

$a(x)=2e^{2x}+24e^{-6x}$

Explanation:

The acceleration function a(x) can be derived from the velocity function v(x) by taking the derivative: a(x)=v'(x) . So we get

$a(x)=v'(x)=\frac{d}{dx}(e^{2x}-4e^{-6x})=2e^{2x}+24e^{-6x}$

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Example Question #47 : First And Second Derivatives Of Functions

Find the derivative of $y= e^{-8x}\cos(x)$

Possible Answers:

$-8\sin(x)$

$-8e^{-8x}\cos(x)$

$\cos(x)e^{-8x}$

$e^{-8x}(-8\cos(x)-\sin(x))$

None of the other answers

Correct answer:

$e^{-8x}(-8\cos(x)-\sin(x))$

Explanation:

We need to use the Product Rule to evaluate the derivative here.

The formula for the Product Rule is

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

If $f(x) = e^{-8x}$ , $g(x)=\cos(x)$ , then $f'(x)=-8e^{-8x}$ , $g'(x)=-\sin(x)$

Plugging these into our formula, we get-

$(fg)'(x) = (e^{-8x})(-\sin(x))+(-8e^{-8x})(\cos(x))$

$= e^{-8x}(-8\cos(x)-\sin(x))$

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Example Question #251 : Derivative Review

Find the derivative of the function $y= \ln(\pi e^{15x})$

Possible Answers:

$\pi x \ln(e^{15x})$

$15+\pi$

$15\pi\ln(e^{15x})$

None of the other answers

Correct answer:

Explanation:

We could use the chain rule for this function, but we can save some time and work if we recognize that the function can be simplified a bit.

$y = \ln(\pi e^{15x})$

$= \ln(\pi) + \ln(e^{15x}})$

$= \ln(\pi) + 15x$

Hence

y' = 0 + 15 = 15

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Example Question #252 : Derivative Review

Evaluate $\frac{d}{dx}(xe^{\sin x})$

Possible Answers:

$(x\sin x +1)e^{\sin x}$

None of the other answers

$(x\cos x+1)e^{\sin x}$

$xe^{\sin x}$

$(xe^{\cos{x}}+1)x$

Correct answer:

$(x\cos x+1)e^{\sin x}$

Explanation:

To evaluate this derivative, we use the Product Rule.

$\frac{d}{dx}(xe^{\sin x})$

$= (x)(\cos xe^{\sin x})+(1)(e^{\sin x})$ . Use the Product Rule. Keep in mind that the derivative of $e^{\sin x}$ involves the Chain Rule.

$=(x\cos x +1)e^{\sin x}$ . Factor out an $e^{\sin x}$ .

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Example Question #253 : Derivative Review

Find the derivative of the following function:

$f(x)=\cos(3xe^{3x})$

Possible Answers:

$-\sin(3xe^{3x})(12e^{3x})$

$\sin(3xe^{3x})(3e^{3x}+9xe^{3x})$

$-\sin(3xe^{3x})(3e^{3x}+9xe^{3x})$

$-\sin(3xe^{3x})(e^{3x}+9xe^{3x})$

Correct answer:

$-\sin(3xe^{3x})(3e^{3x}+9xe^{3x})$

Explanation:

The derivative of the function is

$f'(x)=-\sin(3xe^{3x})(3e^{3x}+9xe^{3x})$

and was found using the following rules:

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

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Example Question #254 : Derivative Review

What is the second derivative of the following function:

$f(x)=ln(lnx)+e^{sinx}$

Possible Answers:

$f''(x)=\frac{1}{xlnx}+cosxe^{sinx}$

$f''(x)=ln(\frac{1}{x})+e^{sinx}$

$f''(x)=\frac{1}{x^{2}ln^{2}x}+e^{sinx}$

$f''(x)=\frac{-1}{x^{2}lnx}-\frac{1}{x^{2}ln^{2}x}+e^{sinx}cos^{2}x-sinxe^{sinx}$

$f''(x)=ln\frac{1}{x}+cosxe^{sinx}$

Correct answer:

$f''(x)=\frac{-1}{x^{2}lnx}-\frac{1}{x^{2}ln^{2}x}+e^{sinx}cos^{2}x-sinxe^{sinx}$

Explanation:

To solce this problem we use the chain rule.

Taking the first derviative we get:

$f'(x)=\frac{1}{lnx}(\frac{1}{x})+e^{sinx}cosx$ , which simplifies to $f'(x)=\frac{1}{xlnx}+cosxe^{sinx}$

To take the second derivative, we use a combination of the chain and product rules. To use the chain rule on the first term of the equation, we can re-write $\frac{1}{xlnx}$ as $(xlnx)^{-1}$ . Taking the second derivative, we get the following:

$f''(x)= -(xlnx)^{-2}[(x)(\frac{1}{x})+(1)lnx)]+[(cosx)e^{sinx}(cosx)-sinxe^{sinx}]$

Which simplifies to

$f''(x)=-\frac{1}{x^{2}ln^{2}x}(lnx+1)+cos^{2}xe^{sinx}-sinxe^{sinx}$

Which simplifies further to:

$f''(x)=\frac{-1}{x^{2}lnx}-\frac{1}{x^{2}ln^{2}x}+cos^{2}xe^{sinx}-sinxe^{sinx}$

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Example Question #51 : First And Second Derivatives Of Functions

Find the first derivative of the function:

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

Possible Answers:

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

$6xe^{3x^2}\sin(e^{3x^2})$

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

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

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

Correct answer:

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

Explanation:

The derivative of the function is equal to

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

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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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #43 : First And Second Derivatives Of Functions

Example Question #241 : Derivatives

Example Question #45 : First And Second Derivatives Of Functions

Example Question #46 : First And Second Derivatives Of Functions

Example Question #47 : First And Second Derivatives Of Functions

Example Question #251 : Derivative Review

Example Question #252 : Derivative Review

Example Question #253 : Derivative Review

Example Question #254 : Derivative Review

Example Question #51 : First And Second Derivatives Of Functions

All Calculus 2 Resources

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