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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 #171 : First And Second Derivatives Of Functions

Find $\large f''(x)$ if f'(x)=2x^3-7x ?

Possible Answers:

-6x^2+7

6x^2-7

6x^2+7

-6x^2-7

Correct answer:

6x^2-7

Explanation:

Step 1: We just take the derivative of f'(x) , which is given to us in the question.

(2x^3-7x)'

Step 2: Use the power rule to take the derivative..

First Term: $2x^3\rightarrow 6x^2$
Second Term: $-7x\rightarrow-7$

Step 3: Take the final answers in Step 2 and add them together...

The second derivative, f''(x) , is 6x^2-7

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Example Question #1501 : Calculus Ii

Find the first derivative of: $\large y=\left [ (2x+1)^2(x+7)^3\right ]$

Possible Answers:

$\large \large \large \large \large 16x^4-266x^3+1428x^2-2267x+833$

$\large 0$

$\large \large 16x^4+266x^3+1428x^2+2276x+833$

$\large -16x^4+266x^3-1428x^2+2276x-833$

Correct answer:

$\large \large 16x^4+266x^3+1428x^2+2276x+833$

Explanation:

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Example Question #1502 : Calculus Ii

Find the first derivative of: $f(x)=3x^3+\frac{1}{2}x^2-4$ .

Possible Answers:

$f{}'(x)=9x^2+3x$

$f{}'(x)=9x^2+x-1$

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

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

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

Correct answer:

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

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:

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

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Example Question #179 : First And Second Derivatives Of Functions

What is the first derivative of $\large \large f(x)=\sqrt {x}$

Possible Answers:

$\large \frac {1}{2\sqrt{x}}$

$\large \frac {1}{2x}$

$\large 0$

$\large \frac {1}{\sqrt {x}}$

Correct answer:

$\large \frac {1}{2\sqrt{x}}$

Explanation:

Step 1: Re-write $\large f(x)$ as $\large x$ to a power:

$\large \sqrt{x}=x^\frac {1}{2}$
Step 2: Use the power rule:

$\large x^\frac {1}{2}=\frac {1}{2}x^{\frac {1}{2}-1}$

Step 3: Simplify the exponent:

$\large \frac {1}{2}x^{\frac {1}{2}}-1=\frac {1}{2}x^\frac {-1}{2}$

Step 4: Re-write $\large x^\frac {-1}{2}$ as a fraction with a positive exponent:

$\large x^\frac {-1}{2}=\frac {1}{x^\frac {1}{2}}=\frac {1}{\sqrt{x}}$
Step 5: Multiply the result in Step 4 with the coefficient $\large \frac {1}{2}$ in step 3:

$\large \frac {1}{2} \cdot \frac {1}{\sqrt{x}}=\frac {1}{2\sqrt{x}}$

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Example Question #1503 : Calculus Ii

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=-11cos(14x^{2})\end{align*}$

Possible Answers:

$11sin(cos(14x^{2}))$

28x

$-11cos(28xsin(14x^{2}))$

$308xsin(14x^{2})$

Correct answer:

$308xsin(14x^{2})$

Explanation:

$\begin{align*}&\text{The function }f(x)=-11cos(14x^{2})\text{, can be seen as a function nested within another.}\\&\text{Notice the }14x^{2}\text{ within the }-11cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&u=14x^{2}\\&du=28x\\&f'(x)=308xsin(14x^{2})\end{align*}$

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Example Question #181 : First And Second Derivatives Of Functions

$\begin{align*}&\text{Calculate the derivative of the function}\\&12cos(14e^{(x)})\end{align*}$

Possible Answers:

$-168e^{(x)}sin(14e^{(x)})$

$12cos(14e^{(x)})$

$14e^{(x)}$

$-12sin(14e^{(x)})$

Correct answer:

$-168e^{(x)}sin(14e^{(x)})$

Explanation:

$\begin{align*}&\text{Our function, }12cos(14e^{(x)})\text{, is one function nested within another.}\\&\text{Note the }14e^{(x)}\text{ inside of the }12cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&u=14e^{(x)}\\&du=14e^{(x)}\\&f'(x)=14e^{(x)}(-12sin(14e^{(x)}))=-168e^{(x)}sin(14e^{(x)})\end{align*}$

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Example Question #182 : First And Second Derivatives Of Functions

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=-9sin(15cos(3x))\end{align*}$

Possible Answers:

405sin(3x)cos(15cos(3x))

9cos(15cos(3x))

45sin(3x)

9sin(45sin(3x))

Correct answer:

405sin(3x)cos(15cos(3x))

Explanation:

$\begin{align*}&\text{The function }f(x)=-9sin(15cos(3x))\text{, can be seen as a function nested within another.}\\&\text{Notice the }-15cos(3x)\text{ within the }9sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&u=-15cos(3x)\\&du=45sin(3x)\\&f'(x)=45sin(3x)(9cos(15cos(3x)))=405sin(3x)cos(15cos(3x))\end{align*}$

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Example Question #1504 : Calculus Ii

$\begin{align*}&\text{Calculate the derivative of the function}\\&13ln(10ln(x))\end{align*}$

Possible Answers:

$\frac{10}{x}$

$\frac{13}{(xln(x))}$

$\frac{13}{(10ln(x))}$

$13ln(\frac{10}{x})$

Correct answer:

$\frac{13}{(xln(x))}$

Explanation:

$\begin{align*}&\text{Our function, }13ln(10ln(x))\text{, is one function nested within another.}\\&\text{Note the }10ln(x)\text{ inside of the }13ln()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[ln(u)]=\frac{du}{u}\\&u=10ln(x)\\&du=\frac{10}{x}\\&f'(x)=\frac{10}{x}(\frac{13}{(10ln(x))})=\frac{13}{(xln(x))}\end{align*}$

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Example Question #184 : First And Second Derivatives Of Functions

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-11tan(12ln(3x))\end{align*}$

Possible Answers:

$-11tan(\frac{12}{x})$

$-\frac{(132\cdot (tan(12ln(3x))^{2} + 1))}{x}$

$-\frac{12}{x}$

$11tan(12ln(3x))^{2} + 11$

Correct answer:

$-\frac{(132\cdot (tan(12ln(3x))^{2} + 1))}{x}$

Explanation:

$\begin{align*}&\text{The function }y=-11tan(12ln(3x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }-12ln(3x)\text{ is inside of the }11tan()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[ln(u)]=\frac{du}{u}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&u=-12ln(3x)\\&du=-\frac{12}{x}\\&\frac{dy}{dx}=-\frac{12}{x}(11tan(12ln(3x))^{2} + 11)=-\frac{(132\cdot (tan(12ln(3x))^{2} + 1))}{x}\end{align*}$

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Example Question #185 : First And Second Derivatives Of Functions

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-5sin(9tan(3x))\end{align*}$

Possible Answers:

$27tan(3x)^{2} + 27$

$-5cos(9tan(3x))\cdot (27tan(3x)^{2} + 27)$

$-5sin(27tan(3x)^{2} + 27)$

-5cos(9tan(3x))

Correct answer:

$-5cos(9tan(3x))\cdot (27tan(3x)^{2} + 27)$

Explanation:

$\begin{align*}&\text{The function }y=-5sin(9tan(3x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }9tan(3x)\text{ is inside of the }-5sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[sin(u)]=cos(u)du\\&u=9tan(3x)\\&du=27tan(3x)^{2} + 27\\&\frac{dy}{dx}=27tan(3x)^{2} + 27(-5cos(9tan(3x)))=-5cos(9tan(3x))\cdot (27tan(3x)^{2} + 27)\end{align*}$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #171 : First And Second Derivatives Of Functions

Example Question #1501 : Calculus Ii

Example Question #1502 : Calculus Ii

Example Question #179 : First And Second Derivatives Of Functions

Example Question #1503 : Calculus Ii

Example Question #181 : First And Second Derivatives Of Functions

Example Question #182 : First And Second Derivatives Of Functions

Example Question #1504 : Calculus Ii

Example Question #184 : First And Second Derivatives Of Functions

Example Question #185 : First And Second Derivatives Of Functions

All Calculus 2 Resources

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