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Calculus 1 : Differentiable Rate

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Example Questions

Example Question #1 : Differentiable Rate

For the relation $\small x^2y+2x-y^2=x$ , compute $\small y'$ using implicit differentiation.

Possible Answers:

$\small \small \small \small y'=\frac{2xy+1}{x^2-2y}$

$\small \small \small \small y'=-\frac{2xy+1}{y(x^2-2y)}$

$\small \small \small \small y'=-\frac{2xy+1}{x^2-2y}$

$\small \small \small \small y'=-\frac{2x+1}{x^2-2y}$

Correct answer:

$\small \small \small \small y'=-\frac{2xy+1}{x^2-2y}$

Explanation:

Computing $\small y'$ of the relation $\small x^2y+2x-y^2=x$ can be done through implicit differentiation:

$\small \small \frac{d}{dx}(x^2y+2x-y^2)=\frac{d}{dx}x$

$\small \small \implies 2xy+x^2y'+2-2yy'=1$

Now we isolate the $\small y'$ :

$\small \small \small \iff x^2y'-2yy'=1-2-2xy=-2xy-1$

$\small \iff y'(x^2-2y)=-(2xy+1)$

$\small \small \implies y'=-\frac{2xy+1}{x^2-2y}$

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Example Question #2 : Differentiable Rate

In chemistry, rate of reaction is related directly to rate constant .

$\frac{dC}{dt}=k$ , where is the initial concentration

Give the concentration of a mixture with rate constant $k=\frac{3}{sec}$ and initial concentration C(0)=40M , seconds after the reaction began.

Possible Answers:

30M

70M

$30e^{(-40)}$

$40e^{(-30)}$

Correct answer:

70M

Explanation:

This is a simple problem of integration. To find the formula for concentration from the formula of concentration rates, you simply take the integral of both sides of the rate equation with respect to time.

$\int \frac{dC}{dt} dt=\int (k)dt$

Therefore, the concentration function is given by

C=kt+C_0 , where C_0 is the initial concentration.

Plugging in our values,

C=3*10+40= 70M

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Example Question #2901 : Calculus

and are related by the function $y=3x^{4}-x+5$ . Find $\frac{dy}{dt}$ at t=2 if x=1 and $\frac{dx}{dt}=6$ at t=2 .

Possible Answers:

$\frac{dy}{dt}=73$

$\frac{dy}{dt}=66$

$\frac{dy}{dt}=6$

$\frac{dy}{dt}=60$

$\frac{dy}{dt}=12$

Correct answer:

$\frac{dy}{dt}=66$

Explanation:

We will use the chain and power rules to differentiate both sides of this equation.

Power Rule:

$\frac{d}{dx}x^n=nx^{n-1}$

Chain Rule:

$\frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$ .

Applying the above rules to our function we find the following derivative.

$\frac{dy}{dt}=\frac{d}{dt}[3x^{4}-x+5]$

$\frac{dy}{dt}=12x^{3}\frac{dx}{dt}-1\frac{dx}{dt}$

at t=2 , $\frac{dx}{dt}=6$ and x=1 .

Therefore at t=2

$\frac{dy}{dt}=12(1)^{3}(6)-1(6)=72-6=66$

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Example Question #4 : How To Find Differentiable Of Rate

Let $y = x^{xlnx-e}.$ Use logarithmic differentiation to find y'(e) .

Possible Answers:

Correct answer:

Explanation:

The form of log differentiation after first "logging" both sides, then taking the derivative is as follows:

$\frac{1}{y}y' = \frac{d}{dx}(lnf(x))$ which implies

$y' = f(x)\frac{d}{dx}(lnf(x))$

So:

$y'(e)=e^{elne-e}(ln^{2}e+2lne-\frac{e}{e})= 1(1+2-1)= 2$

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Example Question #1 : Differentiable Rate

We can interperet a derrivative as $\frac{dy}{dx}= \lim_{\Delta \rightarrow 0} \frac{\Delta y}{\Delta x}$ (i.e. the slope of the secant line cutting the function as the change in x and y approaches zero) but these so-called "differentials" ( $\Delta x$ and $\Delta y$ ) can be a good tool to use for aproximations. If we suppose that $\frac{dy}{dx}=f'(x)$ , or equivalently dy=f'(x)dx . If we suppose a change in x (have a concrete value for $\Delta x$ ) we can find the change in with the afore mentioned relation.

Let f(x)= 3x^2 + x -1 . Find f'(3) and, given $\Delta x = 0.01,$ and find $\Delta y$ .

Possible Answers:

0.1

-0.19

0.034

1.9

0.19

Correct answer:

0.19

Explanation:

Taking the derivative of the function:

$f'(x)= \frac{d}{dx}[3x^2 + x -1]$

$\frac{dy}{dx}= 6x + 1$

Evaluating at x=3 :

$\frac{dy}{dx}|_{x=3} = 6(3) + 1$

$\frac{dy}{dx}|_{x=3} = 6(3) + 1=19$

Manipulating the equation:

dy =19dx

Allowing dx to be .01:

dy=19*0.01

dy=0.19

Which is our answer.

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Example Question #5 : How To Find Differentiable Of Rate

Let $f(x)= \tan (\cos x)$ . Find $f'(\frac{\pi}{4})$ given $\Delta y = 0.3$ , find $\Delta x.$

Possible Answers:

$\approx 0.3422$

$\approx 0.2327$

$\approx -1.0736$

$\approx -0.3669$

$\approx 1.223$

Correct answer:

$\approx -0.3669$

Explanation:

First, we take the derivative of the function:

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

$=\sec^2(\cos x)*(-\sin x)$

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

evaluate the derivative at $x= \frac{\pi}{4}$

$=-\sin (\frac{\pi}{4}) \sec^2(\cos (\frac{\pi}{4}))$

$=-\frac{\sqrt{2}}{2} \sec^2(\frac{\sqrt{2}}{2})$

Manipulating the equation by solving for dy:

$dy=-\frac{\sqrt{2}}{2} \sec^2(\frac{\sqrt{2}}{2})dx$

Assuming dx = 0.3

$dy=-\frac{\sqrt{2}}{2} \sec^2(\frac{\sqrt{2}}{2})*(0.3)$

$\approx -0.3669$

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Example Question #4 : Differentiable Rate

The find the change of volume of a spherical balloon that is growing from to 5.5in.

Possible Answers:

$50\pi\; cubic \;inches$

$5\pi\; cubic \;inches$

$25\pi\; cubic \;inches$

$10\pi\;cubic\;inches$

$100\pi\; cubic \;inches$

Correct answer:

$50\pi\; cubic \;inches$

Explanation:

This is a related rate problem. To find the rate of change of volume with respect to radius, we need to take the derivative of the volume of a sphere equation $(\frac{4}{3}\pi r^{3}).$

Then, we will plug in the relevant information. The initial radius will be substituted in for , and dr=0.5in , since that is the change from the initial to final radius of the balloon.

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

$dV=4\pi (5in)^{2}(0.5in)=50\pi in^{3}.$

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Example Question #5 : How To Find Differentiable Of Rate

Find the rate of change of $y=4x^{3}-12x^{2}+6x+7$ at x=2 .

Possible Answers:

Correct answer:

Explanation:

To find the rate of change of a polynomial at a point, we must find the first derivative of the polynomial and evaluate the derivative at that point.

For this problem,

$y=4x^{3}-12x^{2}+6x+7$

the first derivative of this expression is

$\frac{dy}{dx}=12x^{2}-24x+6$

at x=2 the rate of change is

$\frac{dy}{dx}=12*(2)^{2}-24(2)+6=48-48+6=6$

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Calculus 1 : Differentiable Rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1 : Differentiable Rate

Example Question #2 : Differentiable Rate

Example Question #2901 : Calculus

Example Question #4 : How To Find Differentiable Of Rate

Example Question #1 : Differentiable Rate

Example Question #5 : How To Find Differentiable Of Rate

Example Question #4 : Differentiable Rate

Example Question #5 : How To Find Differentiable Of Rate

All Calculus 1 Resources

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