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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 #302 : Derivative Review

Find the first derivative of the function:

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

Possible Answers:

$\ln(x)+1+e^{2x}$

$\ln(x)+2e^{2x}$

$\ln(x)+x+2e^{2x}$

$\ln(x)+1+2e^{2x}$

Correct answer:

$\ln(x)+1+2e^{2x}$

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

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

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Example Question #103 : First And Second Derivatives Of Functions

Find the second derivative of f(x)=sinx+tanx.

Possible Answers:

f''(x)=secxtanx+sinx

f''(x)=2sec^2xtanx-sinx

f''(x)=cos(x)

f''(x)=2sec^2xtanx+sinx

f''(x)=secxtanx-sinx

Correct answer:

f''(x)=2sec^2xtanx-sinx

Explanation:

To find the second derivative of a function, we must first find the first derivative.

f(x)=sinx+tanx.

f'(x)=cosx+sec^2x. Now, we must differentiate this function. The first part is a straightforward trigonometric derivative. For the second term, we must use the chain rule.

$f''(x)=-sinx+2secx*\frac{d(secx)}{dx}$

=-sinx+2secx*secxtanx=-sinx+2sec^2xtanx.

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Example Question #104 : First And Second Derivatives Of Functions

Find the derivative of $e^{3x^2}$ .

Possible Answers:

$6x*e^{3x^2}$

$ln(6x)*e^{3x^2}$

$3x^2*e^{3x^2}$

$ln(3x^2)*e^{3x^2}$

$e^{6x}$

Correct answer:

$6x*e^{3x^2}$

Explanation:

The derivative of e^x is special because it returns back the original function (e^x) .

Therefore, this derivative is simply a chain rule application of that exponential function.

Recall the chain rule is,

$\frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$

Applying this rule to the original function results in the following derivative.

$\frac{d}{dx}e^{3x^2}=e^u\frac{du}{dx}=e^{3x^2}*6x.$

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Example Question #105 : First And Second Derivatives Of Functions

Find the second derivative of f(x)=sinx-cos(x).

Possible Answers:

sin^2(x)

tan(x)

sinx+cosx

sinx-cosx

cosx-sinx

Correct answer:

cosx-sinx

Explanation:

To get to the second derivative, we must first take the first derivative!

f(x)=sinx-cosx.

f'(x)=cosx--sinx=cosx+sinx.

Then, we will differentiate that to get our answer.

f''(x)=-sinx+cosx.

Recall the trigonometric derivatives:

$\frac{d}{dx}sin(x)=cos(x); \frac{d}{dx}cos(x)=-sin(x)$

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Example Question #1431 : Calculus Ii

What is the derivative of f(x)=9x^2-5x+1 ?

Possible Answers:

f{}'(x)=18x

f{}'(x)=18x-5

f{}'(x)=x-5

$f{}'(x)=18x-5x^2$

f{}'(x)=18

Correct answer:

f{}'(x)=18x-5

Explanation:

When taking the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent. Therefore, the answer is: f{}'(x)=18x-5.

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Example Question #311 : Derivative Review

What is the derivative of $f(x)=5x^{\frac{4}{5}}-2x$ ?

Possible Answers:

$f{}'(x)=\frac{4}{x^{\frac{1}{5}}}-2x$

$f{}'(x)=\frac{2}{x^{\frac{1}{5}}}-2$

$f{}'(x)=\frac{4}{x^{\frac{1}{5}}}-2$

$f{}'(x)=\frac{4}{x^-{\frac{1}{5}}}-2$

$f{}'(x)=\frac{4}{x^{\frac{1}{5}}}$

Correct answer:

$f{}'(x)=\frac{4}{x^{\frac{1}{5}}}-2$

Explanation:

Remember that when taking the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent: $5(\frac{4}{5})x^{\frac{-1}{5}}-2$ . Simplify and make sure you don't have a negative exponent (to make it positive, put it in the denominator). Therefore, your answer is $f{}'(x)=\frac{4}{x^{\frac{1}{5}}}-2$ .

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Example Question #112 : First And Second Derivatives Of Functions

What is the derivative of $f(x)=9x^{\frac{5}{4}}-3x?$

Possible Answers:

$f{}'(x)=\frac{45}{4}x^{\frac{1}{4}}-3$

$f{}'(x)=\frac{45}{2}x^{\frac{1}{4}}-3$

$f{}'(x)=\frac{45}{4}x^{\frac{1}{4}}-3x$

$f{}'(x)=\frac{45}{4}x^{\frac{3}{4}}-3$

$f{}'(x)=\frac{45}{4}x^{\frac{1}{4}}-5$

Correct answer:

$f{}'(x)=\frac{45}{4}x^{\frac{1}{4}}-3$

Explanation:

Remember that when taking the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent.

Therefore, your answer should look like this:

$f{}'(x)=\frac{45}{4}x^{\frac{1}{4}}-3$

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Example Question #111 : First And Second Derivatives Of Functions

What is the derivative of f(x)=3x^3-2x^2+4x-1?

Possible Answers:

f{}'(x)=18x-4

$f{}'(x)=9x^2-4x+4$

$f{}'(x)=9x^2-4x$

f{}'(x)=9x+4

Correct answer:

$f{}'(x)=9x^2-4x+4$

Explanation:

Recall that when taking the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent.

Therefore, after differentiating, you get

$f'(x)=3\cdot 3x^{3-1}-2\cdot 2x^{2-1}+1\cdot 4x^{1-1}$

$f{}'(x)=9x^2-4x+4$ .

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Example Question #114 : First And Second Derivatives Of Functions

Find the derivative of the function:

$f(x)=x^3+e^x\sin(x)+e^{4x}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

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

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Example Question #112 : First And Second Derivatives Of Functions

Find the second derivative of the function

$f(x)=x^2\sin(2x)$

Possible Answers:

$(2+4x)(\sin(2x))+8x(\cos(2x))$

$(2+4x^2)(\sin(2x))+8x(\cos(2x))$

$(2-4x^2)(\sin(2x))+8x\cos(2x)$

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

Correct answer:

$(2-4x^2)(\sin(2x))+8x\cos(2x)$

Explanation:

To begin, you must take the first derivative of the function, which is:

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

Then, you take the derivative of the first derivative, which equals:

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

This simplifies to:

$(2-4x^2)(\sin(2x))+8x(\cos(2x))$

The rules of differentiation that were used are:

$\frac{d}{dx}(f(x)g(x))=f'(x)g(x)+f(x)g'(x)$

$\frac{d}{dx}f(g(x))=f'(g(x))(g'(x))$

$\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}(f(x))+\frac{d}{dx}(g(x))$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #302 : Derivative Review

Example Question #103 : First And Second Derivatives Of Functions

Example Question #104 : First And Second Derivatives Of Functions

Example Question #105 : First And Second Derivatives Of Functions

Example Question #1431 : Calculus Ii

Example Question #311 : Derivative Review

Example Question #112 : First And Second Derivatives Of Functions

Example Question #111 : First And Second Derivatives Of Functions

Example Question #114 : First And Second Derivatives Of Functions

Example Question #112 : First And Second Derivatives Of Functions

All Calculus 2 Resources

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