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Calculus 1 : Calculus

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Example Questions

Example Question #3071 : Calculus

Use normal rules of differentiation to find the derrivative of the function $f(x)=8x^2\ln x$ .

Possible Answers:

$16x\ln x -8x$

$4x(2 \ln x+1)$

$24\ln x(\frac{1}{x} +1)$

$8x(2\ln x +1)$

Correct answer:

$8x(2\ln x +1)$

Explanation:

We notice that this is the product of two functions, so we'll use the product rule.

$f'(x)=\frac{d}{dx}[8x^2\ln x]$

$=\frac{d}{dx}[8x^2] \ln x+8x^2 \frac{d}{dx}[\ln x]$

simplifying:

$=16x \ln x+8x^2 (\frac{1}{x})$

simplying:

$=16x \ln x+8x$

factoring out , we have

$=8x(2 \ln x+1)$

which is our answer.

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Example Question #152 : How To Find Rate Of Change

Use normal rules of differentiation to find the derrivative of the function $f(x)=\frac{\arctan x}{\tan x}$ .

Possible Answers:

$\frac{\cos^2x}{1+x^2}}$

This function is not differentiable at any point on its domain.

$\frac{\cos^2x- \arctan x}{\tan x (1+x^2)}$

$\frac{\sec x(\frac{1}{1+x^2})-sec^2x}{\tan^2x}$

$\cot x \csc x (\frac{\sin x}{1+x^2}-\arctan x \sec x)$

Correct answer:

$\cot x \csc x (\frac{\sin x}{1+x^2}-\arctan x \sec x)$

Explanation:

We take the derrivative of the function:

$f'(x)=\frac{d}{dx}[\frac{\arctan x}{\tan x}]$

We notice that this is the quotient of two functions, so we apply the quotient rule:

$=\frac{\frac{d}{dx}[\arctan x]\tan x -\arctan x\frac{d}{dx}{[\tan x}]}{\tan^2 x}$

$=\cot^2 x(\frac{1}{1+x^2}\tan x -\arctan x \sec^2 x)$

Factoring out a $\sec x$

$=\cot^2 x \sec x(\frac{\sin x}{1+x^2}-\arctan x \sec x)$ .

Simplifying, we have

$=\cot x \csc x(\frac{\sin x}{1+x^2}-\arctan x \sec x)$

which is our answer.

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Example Question #2041 : Functions

Use normal rules of differentiation to find the derrivative of the function

$f(x)=\frac{\arcsin x}{e^x}$ .

Possible Answers:

$e^{x}(\arcsin x-\frac{1}{\sqrt{1-x^2}})$

$-e^x \arcsin x$

$e^{-x}(\frac{1}{\sqrt{1-x^2}}-\arcsin x)$

$\frac{1}{e^x}(\frac{1}{\sqrt{1-x^2}})$

This function is not differentiable at any point in its domain.

Correct answer:

$e^{-x}(\frac{1}{\sqrt{1-x^2}}-\arcsin x)$

Explanation:

We take the derrivative

$f'(x)=\frac{d}{dx}[\frac{\arcsin x}{e^x}]$

This function is the quotient of two functions, so we use the quotient rule:

$= \frac{\frac{d}{dx}[\arcsin x] e^x-\arcsin x\frac{d}{dx}[e^x]}{(e^x)^2}$

$= \frac{\frac{1}{\sqrt{1-x^2}} e^x-\arcsin x e^x}{(e^{2x})}$

Simplifying, we have

$=e^{-x}(\frac{1}{\sqrt{1-x^2}}-\arcsin x )$

which is our answer.

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Example Question #161 : Rate Of Change

Say $\Delta x$ signifies the change in within the interval [x,x+d] ,

and $\Delta y$ signifies the change on some interval [f(x), f(x+d)] .

In this problem we will use the definition of the derrivative

$\frac{df}{dx}=\lim_{\Delta \rightarrow 0} \frac{\Delta y}{\Delta x}=\lim_{d \rightarrow 0} \frac{f(x+d)-f(x)}{d}$

to compute the instantanious rate of change of the function.

Compute the instantaneous rate of change at f(7) when $f(x)=\frac{1}{2} x+3$ using the definition of the derrivative.

$\lim_{d \rightarrow 0} \frac{f(x+d)-f(x)}{d}$ .

Possible Answers:

$\frac{13}{2}$

$-\frac{1}{2}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

f'(x)= $\lim_{d \rightarrow 0} \frac{f(x+d)-f(x)}{d}$

substituting the function in, we have:

$=\lim_{d \rightarrow 0}\frac{[\frac{1}{2}(x+d)+3]-[\frac{1}{2}x+3]}{d}$

simplifying:

$=\lim_{d \rightarrow 0}\frac{\frac{1}{2}x+\frac{1}{2}d+3-\frac{1}{2}x-3}{d}$

canceling terms:

$=\lim_{d \rightarrow 0}\frac{\frac{1}{2}d}{d}$

and canceling terms, we have that:

$=\lim_{d \rightarrow 0} \frac{1}{2}$

$=\frac{1}{2}$

Notice that all the 's and 's all canceled out. This always happens with linear functions. We conclude that the instantaneous rate of change for all values of x is $\frac{1}{2}$ including when x is. Hence,

$f'(7)=\frac{1}{2}$ .

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Example Question #162 : Rate Of Change

Say $\Delta x$ signifies the change in within the interval [x,x+d] ,

and $\Delta y$ signifies the change on some interval [f(x), f(x+d)] .

In this problem we will use the definition of the derrivative

$\frac{df}{dx}=\lim_{\Delta \rightarrow 0} \frac{\Delta y}{\Delta x}=\lim_{d \rightarrow 0} \frac{f(x+d)-f(x)}{d}$

to compute the instantanious rate of change of the function.

Compute the instantaneous rate of change at f(7) when f(x)=2x^2-3x+7 using the definition of a derivative: $f'(x)=\lim_{d \rightarrow 0}\frac{f(x+d)-f(x)}{d}$ .

Possible Answers:

Correct answer:

Explanation:

using $f'(x)=\lim_{d \rightarrow 0}\frac{f(x+d)-f(x)}{d}$ we have:

$f'(x)=\lim_{d \rightarrow 0}\frac{[2(x+d)^2-3(x+d)+7]-[2x^2-3x+7]}{d}$

simplifying, we have:

$=\lim_{d \rightarrow 0}\frac{2x^2+4xd+2d^2-3x-3d+7-2x^2+3x-7}{d}$

Canceling out terms:

$=\lim_{d \rightarrow 0}\frac{4xd+2d^2-3d}{d}$

Factoring out a d:

$=\lim_{d \rightarrow 0} \frac{d(4x+2d-3)}{d}$

canceling out the d's

$=\lim_{d \rightarrow 0} 4x+2d-3$

applying the limit:

= 4x+0-3 = 4x-3

So, to find the instantaneous rate of change of f(7) we just plug in as the input:

4(7)-3=25 which is our answer.

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Example Question #163 : Rate Of Change

A cube is diminishing in size. What is the length of the diagonal of the cube at the time that the rate of shrinkage of the cube's diagonal is equal to twice the rate of shrinkage of its surface area?

Possible Answers:

$\frac{1}{4}$

$\frac{1}{8}$

$\frac{\sqrt{3}}{2}$

$\frac{\sqrt{3}}{4}$

$\frac{\sqrt{3}}{8}$

Correct answer:

$\frac{1}{8}$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its diagonal and surface area in terms of the length of its sides:

$d=\sqrt{3s^2}$

$d=s\sqrt{3}$

A=6s^2

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

$\frac{ds}{dt}=\sqrt{3}\frac{ds}{dt}$

$\frac{dA}{dt}=12s\frac{ds}{dt}$

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of shrinkage of the cube's diagonal is equal to twice the rate of shrinkage of its surface area:

$\sqrt{3}\frac{ds}{dt}=(2)12s\frac{ds}{dt}$

$s=\frac{\sqrt{3}}{24}$

Again, the diagonal of a cube is given by the equation:

$d=\sqrt{3s^2}$

$d=s\sqrt{3}$

$d=\frac{\sqrt{3}}{24}\sqrt{3}$

$d=\frac{1}{8}$

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Example Question #2042 : Functions

A cube is diminishing in size. What is the length of the sides of the cube at the time that the rate of shrinkage of the cube's surface area is equal to a third of the rate of shrinkage of its volume?

Possible Answers:

108

Correct answer:

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its volume and surface area in terms of the length of its sides:

V=s^3

A=6s^2

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

$\frac{dA}{dt}=12s\frac{ds}{dt}$

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of shrinkage of the cube's surface area is equal to a third of the rate of shrinkage of its volume:

$\frac{1}{3}3s^2\frac{ds}{dt}=12s\frac{ds}{dt}$

s^2=12s

s=12

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Example Question #162 : How To Find Rate Of Change

A spherical balloon is deflating, and somehow still maintaining its spherical shape. Madness. What is the volume of the sphere at the instance the rate of shrinkage of the volume is twice times the rate of shrinkage of the radius?

Possible Answers:

$\sqrt{\frac{9}{2\pi}}$

$\sqrt{\frac{2\pi}{9}}$

$\frac{2\pi}{9}$

$\sqrt{\frac{2}{9\pi}}$

$\pi \sqrt{\frac{2}{9}}$

Correct answer:

$\sqrt{\frac{2}{9\pi}}$

Explanation:

Let's begin by writing the equations for the volume of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of shrinkage of the volume is twice times the rate of shrinkage of the radius,let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=2 \frac{dr}{dt}$

$4\pi r^2=2$

$r^2 =\frac{1}{2\pi}$

$r=\sqrt{\frac{1}{2\pi}}$

$V=\frac{4}{3}\pi r^3$

$V=\frac{4}{3}\pi \frac{1}{2\pi}\sqrt{\frac{1}{2\pi}}$

$V=\frac{2}{3}\sqrt{\frac{1}{2\pi}}$

$V=\sqrt{\frac{2}{9\pi}}$

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Example Question #2043 : Functions

A cube is diminishing in size. What is the length of the diagonal of the cube at the time that the rate of shrinkage of the cube's volume is equal to thirty-six times the rate of shrinkage of its sides?

Possible Answers:

$4\sqrt{3}$

$6\sqrt{3}$

$12\sqrt{3}$

Correct answer:

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its volume in terms of the length of its sides:

V=s^3

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of shrinkage of the cube's volume is equal to thirty-six times the rate of shrinkage of its sides:

$3s^2\frac{ds}{dt}=36\frac{ds}{dt}$

s^2=12

$s=\sqrt{12}$

$s=4\sqrt{3}$

The diagonal of a cube is given by the equation:

$d=\sqrt{3s^2}$

$d=4\sqrt{3(}\sqrt{3})$

d=12

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Example Question #161 : Rate Of Change

A cube is diminishing in size. What is the length of the diagonal of the cube at the time that the rate of shrinkage of the cube's sides is equal to six times the rate of shrinkage of its surface area?

Possible Answers:

$\frac{\sqrt{3}}{72}$

$24\sqrt{3}$

$72\sqrt{3}$

Correct answer:

$\frac{\sqrt{3}}{72}$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

A=6s^2

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

$\frac{dA}{dt}=12s\frac{ds}{dt}$

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of shrinkage of the cube's sides is equal to six times the rate of shrinkage of its surface area:

$\frac{ds}{dt}=(6)12s\frac{ds}{dt}$

$s=\frac{1}{72}$

The diagonal of a cube is given by the equation:

$d=\sqrt{3s^2}$

$d=s\sqrt{3}$

$d=\frac{\sqrt{3}}{72}$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3071 : Calculus

Example Question #152 : How To Find Rate Of Change

Example Question #2041 : Functions

Example Question #161 : Rate Of Change

Example Question #162 : Rate Of Change

Example Question #163 : Rate Of Change

Example Question #2042 : Functions

Example Question #162 : How To Find Rate Of Change

Example Question #2043 : Functions

Example Question #161 : Rate Of Change

All Calculus 1 Resources

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