How to find differentiable of rate - AP Calculus AB

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Question

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

Answer

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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Question

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 , seconds after the reaction began.

Answer

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

, where is the initial concentration.

Plugging in our values,

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Question

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

Answer

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 , $\frac{dx}{dt}=6$ and .

Therefore at

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

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Question

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

Answer

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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Question

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 . 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 . Find and, given $\Delta x = 0.01,$ and find $\Delta y$ .

Answer

Taking the derivative of the function:

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

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

Evaluating at :

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

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

Manipulating the equation:

Allowing dx to be .01:

Which is our answer.

Compare your answer with the correct one above

Question

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

Answer

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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Question

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

Answer

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 , 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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Question

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

Answer

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 the rate of change is

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

Compare your answer with the correct one above

Question

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

Answer

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 $\small y'$ :

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

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

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

Compare your answer with the correct one above

Question

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 , seconds after the reaction began.

Answer

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

, where is the initial concentration.

Plugging in our values,

Compare your answer with the correct one above

Question

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

Answer

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 , $\frac{dx}{dt}=6$ and .

Therefore at

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

Compare your answer with the correct one above

Question

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

Answer

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:

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

Compare your answer with the correct one above

Question

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 . 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 . Find and, given $\Delta x = 0.01,$ and find $\Delta y$ .

Answer

Taking the derivative of the function:

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

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

Evaluating at :

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

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

Manipulating the equation:

Allowing dx to be .01:

Which is our answer.

Compare your answer with the correct one above

Question

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

Answer

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

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

$\approx -0.3669$

Compare your answer with the correct one above

Question

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

Answer

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 , since that is the change from the initial to final radius of the balloon.

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

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

Compare your answer with the correct one above

Question

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

Answer

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 the rate of change is

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

Compare your answer with the correct one above

Question

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

Answer

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 $\small y'$ :

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

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

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

Compare your answer with the correct one above

Question

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 , seconds after the reaction began.

Answer

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

, where is the initial concentration.

Plugging in our values,

Compare your answer with the correct one above

Question

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

Answer

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 , $\frac{dx}{dt}=6$ and .

Therefore at

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

Compare your answer with the correct one above

Question

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

Answer

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:

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

Compare your answer with the correct one above

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