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Calculus 2 : Derivatives

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Example Questions

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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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 #1543 : Calculus Ii

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

Possible Answers:

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

-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 : Derivatives

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

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

Example Question #1541 : Calculus Ii

Example Question #1542 : Calculus Ii

Example Question #1543 : Calculus Ii

Example Question #415 : Derivative Review

Example Question #416 : Derivative Review

All Calculus 2 Resources

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