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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #211 : First And Second Derivatives Of Functions

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:

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

g'=f(t)

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

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

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 #411 : Derivative Review

Find the derivative of the function:

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

Possible Answers:

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

$2x\cos(x)$

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

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

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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Example Question #1541 : Calculus Ii

$\begin{align*}&\text{Calculate the derivative of the function}\\&-10tan(20tan(6x))\end{align*}$

Possible Answers:

$- 10tan(20tan(6x))^{2} - 10$

$-10\cdot (120tan(6x)^{2} + 120)\cdot (tan(20tan(6x))^{2} + 1)$

$-10tan(120tan(6x)^{2} + 120)$

$120tan(6x)^{2} + 120$

Correct answer:

$-10\cdot (120tan(6x)^{2} + 120)\cdot (tan(20tan(6x))^{2} + 1)$

Explanation:

$\begin{align*}&\text{Our function, }-10tan(20tan(6x))\text{, is one function nested within another.}\\&\text{Note the }20tan(6x)\text{ inside of the }-10tan()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&u=20tan(6x)\\&du=120tan(6x)^{2} + 120\\&f'(x)=(120tan(6x)^{2} + 120)(- 10tan(20tan(6x))^{2} - 10)=-10\cdot (120tan(6x)^{2} + 120)\cdot (tan(20tan(6x))^{2} + 1)\end{align*}$

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Example Question #1542 : Calculus Ii

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-17e^{(-19\cdot 5^{(4x)})}\end{align*}$

Possible Answers:

$-17e^{(-19\cdot 5^{(4x)})}$

$-17e^{(-76\cdot 5^{(4x)}ln(5))}$

$1292\cdot 5^{(4x)}e^{(-19\cdot 5^{(4x)})}ln(5)$

$-76\cdot 5^{(4x)}ln(5)$

Correct answer:

$1292\cdot 5^{(4x)}e^{(-19\cdot 5^{(4x)})}ln(5)$

Explanation:

$\begin{align*}&\text{The function }y=-17e^{(-19\cdot 5^{(4x)})}\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }-19\cdot 5^{(4x)}\text{ is inside of the }-17e^{()}\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[a^u]=a^uduln(a)\\&d[e^u]=e^udu\\&u=-19\cdot 5^{(4x)}\\&du=-76\cdot 5^{(4x)}ln(5)\\&\frac{dy}{dx}=(-76\cdot 5^{(4x)}ln(5))(-17e^{(-19\cdot 5^{(4x)})})=1292\cdot 5^{(4x)}e^{(-19\cdot 5^{(4x)})}ln(5)\end{align*}$

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Example Question #1541 : Calculus Ii

$\begin{align*}&\text{Calculate the derivative of the function}\\&4^{(15cos(4x))}\end{align*}$

Possible Answers:

-60sin(4x)

$\frac{1}{4^{(60sin(4x))}}$

$-60\cdot 4^{(15cos(4x))}sin(4x)ln(4)$

$\frac{81}{4^{(60sin(4x))}}$

Correct answer:

$-60\cdot 4^{(15cos(4x))}sin(4x)ln(4)$

Explanation:

$\begin{align*}&\text{Our function, }4^{(15cos(4x))}\text{, is one function nested within another.}\\&\text{Note the }15cos(4x)\text{ inside of the }1\cdot4^{()}\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&d[a^u]=a^uduln(a)\\&u=15cos(4x)\\&du=-60sin(4x)\\&f'(x)=(-60sin(4x))(4^{(15cos(4x))}ln(4))=-60\cdot 4^{(15cos(4x))}sin(4x)ln(4)\end{align*}$

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Example Question #415 : Derivative Review

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=\frac{1}{5^{(17x^{2})}}\end{align*}$

Possible Answers:

-34x

$-\frac{(34xln(5))}{5^{(17x^{2})}}$

$\frac{1}{5^{(34x)}}$

$-\frac{(68xln(5))}{5^{(17x^{2})}}$

Correct answer:

$-\frac{(34xln(5))}{5^{(17x^{2})}}$

Explanation:

$\begin{align*}&\text{The function }f(x)=\frac{1}{5^{(17x^{2})}}\text{, can be seen as a function nested within another.}\\&\text{Notice the }-17x^{2}\text{ within the }1\cdot5^{()}\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[a^u]=a^uduln(a)\\&u=-17x^{2}\\&du=-34x\\&f'(x)=(-34x)(\frac{ln(5)}{5^{(17x^{2})}})=-\frac{(34xln(5))}{5^{(17x^{2})}}\end{align*}$

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Example Question #416 : Derivative Review

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-13ln(16cos(3x))\end{align*}$

Possible Answers:

$\frac{(39sin(3x))}{cos(3x)}$

-13ln(-48sin(3x))

$-\frac{13}{(16cos(3x))}$

-48sin(3x)

Correct answer:

$\frac{(39sin(3x))}{cos(3x)}$

Explanation:

$\begin{align*}&\text{The function }y=-13ln(16cos(3x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }16cos(3x)\text{ is inside of the }-13ln()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&d[ln(u)]=\frac{du}{u}\\&u=16cos(3x)\\&du=-48sin(3x)\\&\frac{dy}{dx}=(-48sin(3x))(-\frac{13}{(16cos(3x))})=\frac{(39sin(3x))}{cos(3x)}\end{align*}$

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #211 : First And Second Derivatives Of Functions

Example Question #412 : Derivative Review

Example Question #411 : Derivative Review

Example Question #414 : Derivative Review

Example Question #211 : First And Second Derivatives Of Functions

Example Question #1541 : Calculus Ii

Example Question #1542 : Calculus Ii

Example Question #1541 : Calculus Ii

Example Question #415 : Derivative Review

Example Question #416 : Derivative Review

All Calculus 2 Resources

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