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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 #404 : Derivative Review

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-cos(10tan(5x))\end{align*}$

Possible Answers:

$50tan(5x)^{2} + 50$

$-cos(50tan(5x)^{2} + 50)$

$sin(10tan(5x))\cdot (50tan(5x)^{2} + 50)$

sin(10tan(5x))

Correct answer:

$sin(10tan(5x))\cdot (50tan(5x)^{2} + 50)$

Explanation:

$\begin{align*}&\text{The function }y=-cos(10tan(5x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }10tan(5x)\text{ is inside of the }-1cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[cos(u)]=-sin(u)du\\&u=10tan(5x)\\&du=50tan(5x)^{2} + 50\\&\frac{dy}{dx}=(50tan(5x)^{2} + 50)(sin(10tan(5x)))=sin(10tan(5x))\cdot (50tan(5x)^{2} + 50)\end{align*}$

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Example Question #204 : First And Second Derivatives Of Functions

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=4cos(7sin(x))\end{align*}$

Possible Answers:

-28sin(7sin(x))cos(x)

-7cos(x)

4sin(7sin(x))

4cos(7cos(x))

Correct answer:

-28sin(7sin(x))cos(x)

Explanation:

$\begin{align*}&\text{The function }y=4cos(7sin(x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }-7sin(x)\text{ is inside of the }4cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&u=-7sin(x)\\&du=-7cos(x)\\&\frac{dy}{dx}=(-7cos(x))(4sin(7sin(x)))=-28sin(7sin(x))cos(x)\end{align*}$

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Example Question #406 : Derivative Review

$\begin{align*}&\text{Calculate the derivative of the function}\\&10cos(2\cdot 3^x)\end{align*}$

Possible Answers:

$-20\cdot 3^xln(3)sin(2\cdot 3^x)$

$-2\cdot 3^xln(3)$

$10sin(2\cdot 3^x)$

$10cos(2\cdot 3^xln(3))$

Correct answer:

$-20\cdot 3^xln(3)sin(2\cdot 3^x)$

Explanation:

$\begin{align*}&\text{Our function, }10cos(2\cdot 3^x)\text{, is one function nested within another.}\\&\text{Note the }-2\cdot 3^x\text{ inside of the }10cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&u=-2\cdot 3^x\\&du=-2\cdot 3^xln(3)\\&f'(x)=(-2\cdot 3^xln(3))(10sin(2\cdot 3^x))=-20\cdot 3^xln(3)sin(2\cdot 3^x)\end{align*}$

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Example Question #407 : Derivative Review

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=5sin(3e^{(6x)})\end{align*}$

Possible Answers:

$5sin(18e^{(6x)})$

$90e^{(6x)}cos(3e^{(6x)})$

$5cos(3e^{(6x)})$

$18e^{(6x)}$

Correct answer:

$90e^{(6x)}cos(3e^{(6x)})$

Explanation:

$\begin{align*}&\text{The function }y=5sin(3e^{(6x)})\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }3e^{(6x)}\text{ is inside of the }5sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&u=3e^{(6x)}\\&du=18e^{(6x)}\\&\frac{dy}{dx}=(18e^{(6x)})(5cos(3e^{(6x)}))=90e^{(6x)}cos(3e^{(6x)})\end{align*}$

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Example Question #403 : Derivatives

The Laplace Transform is an integral transform that converts functions from the time domain to the complex frequency domain . The transformation of a function f(t) into its Laplace Transform F(s) is given by:

$F(s)= \int_{0}^{\infty}e^{-st}*f(t) dt$

Where s=a+bi , where and are constants and is the imaginary number.

Determine the value of $\frac{\mathrm{d} }{\mathrm{d} s}F(s)$ .

Possible Answers:

-tF(s)

-sF(s)

tF(s)

f'(t)*F(s)

Correct answer:

-tF(s)

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} s}F(s)= \frac{d}{ds}\int_{0}^{\infty}e^{-st}*f(t) dt=-t*[\int_{0}^{\infty}e^{-st}*f(t) dt]=-tF(s)$

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Example Question #411 : Derivative Review

In exponentially decaying systems, often times the solutions to differential equations take on the form of an integral called Duhamel's Integral. This is given by:

$g(t)=\int_{0}^{t}e^{-s(t-b)}*f(b) db$

Where is a constant and f(b) is a function that represents an external force.

Taking one derivative with respect to , determine which of the following differential equations satisfies.

Possible Answers:

$g'=f(t)-f(0)e^{-st}$

$g'=f(t)-e^{-t}$

g'=f(t)

$g'=f(t)+f(0)e^{-st}$

Correct answer:

$g'=f(t)-f(0)e^{-st}$

Explanation:

Taking one time derivative we get:

$g'(t)=\frac{d}{dt}\int_{o}^{t}e^{-s(t-b)}f(b)dt=e^{-s(t-t)}f(t)-e^{-s(t-0)}f(0)=f(t)-f(0)e^{-st}$

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Example Question #412 : Derivative Review

Find the derivative of the function:

$f(x)=\cos(2x)\tan(2x)$

Possible Answers:

$2\sin(2x)\tan(2x)+\frac{\sec(2x)}{2}$

$-2\sin(2x)\tan(2x)+2\cos(2x)\sec^2(2x)$

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

$-2\sin(2x)\tan(2x)+\sec(2x)$

Correct answer:

$-2\sin(2x)\tan(2x)+2\cos(2x)\sec^2(2x)$

Explanation:

The derivative of the function is equal to

$-2\sin(2x)\tan(2x)+2\cos(2x)\sec^2(2x)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #413 : Derivative Review

Find the derivative of the function:

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

Possible Answers:

$2x\cos(x)$

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

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

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

Correct answer:

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

Explanation:

The derivative of the function is equal to

$f'(x)=2x\cos(x)-x^2\sin(x)$

and was found using the following rules:

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

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Example Question #414 : Derivative Review

Find the absolute maxima of the following function on the given interval:

f(x)=2x^3-6x-1 on the interval (-2,0)

Possible Answers:

(-0.168,0)

(3,1)

(-6,2)

(-1,3)

(0,-1)

Correct answer:

(-1,3)

Explanation:

To find the absolute extrema of a function on a closed interval, one must first take the first derivative of the function.

The derviatve of this function by the power rule is as follows:

f'(x)=6x^2-6

The relative extrema is when the first derivative is equal to 0, that is, there is a change in slope.

Solving for x when it is equal to zero derives:

0=6(x^2-1)=(x+1)(x-1)

Diving by 6 and factoring gives (x+1)(x-1) or x==+1,-1; however, since we are concerned with the interval (-2,0) our x value is -1.

We now however must find the value of f(x) at -1

f(-1)=3

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Example Question #211 : First And Second Derivatives Of Functions

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=2sin(19cos(6x))\end{align*}$

Possible Answers:

-114sin(6x)

2cos(19cos(6x))

-228sin(6x)cos(19cos(6x))

-2sin(114sin(6x))

Correct answer:

-228sin(6x)cos(19cos(6x))

Explanation:

$\begin{align*}&\text{The function }f(x)=2sin(19cos(6x))\text{, can be seen as a function nested within another.}\\&\text{Notice the }19cos(6x)\text{ within the }2sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&u=19cos(6x)\\&du=-114sin(6x)\\&f'(x)=(-114sin(6x))(2cos(19cos(6x)))=-228sin(6x)cos(19cos(6x))\end{align*}$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #404 : Derivative Review

Example Question #204 : First And Second Derivatives Of Functions

Example Question #406 : Derivative Review

Example Question #407 : Derivative Review

Example Question #403 : Derivatives

Example Question #411 : Derivative Review

Example Question #412 : Derivative Review

Example Question #413 : Derivative Review

Example Question #414 : Derivative Review

Example Question #211 : First And Second Derivatives Of Functions

All Calculus 2 Resources

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