Computation of Derivatives - AP Calculus BC

Card 0 of 790

Question

Find $\frac{dy}{dx}$ if $x^3 -x\ln y=ye^x$ .

Answer

This function is implicit, because y is not defined directly in terms of only x. We could try to solve for y, but that would be difficult, if not impossible. The easier solution would be to employ implicit differentiation. Our strategy will be to differentiate the left and right sides by x, apply the rules of differentiation (such as Chain and Product Rules), group dy/dx terms, and solve for dy/dx in terms of both x and y.

$x^3-x\ln y=ye^{x}$

$\frac{d}{dx}[x^3-x\ln y]=\frac{d}{dx}[ye^x]$

$\frac{d}{dx}[x^3]-\frac{d}{dx}[x\ln y]=\frac{d}{dx}[ye^x]$

We will need to apply both the Product Rule and Chain Rule to both the xlny and the terms.

According to the Product Rule, if f(x) and g(x) are functions, then $\frac{d}{dx}[f(x)g(x)]=f(x)\cdot \frac{d}{dx}[g(x)]+g(x)\cdot \frac{d}{dx}[f(x)]$ .

And according to the Chain Rule,

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

$3x^2-(x\cdot \frac{d}{dx}[\ln y]+\ln y\cdot \frac{d}{dx}[x])=e^x\cdot \frac{d}{dx}[y]+y\cdot \frac{d}{dx}[e^x]$

$3x^2-(x\cdot \frac{d}{dy}[\ln y]\cdot \frac{dy}{dx}+\ln y)=e^x\cdot \frac{d}{dy}[y]\cdot \frac{dy}{dx}+y\cdot e^x$

$3x^2-(\frac{x}{y}\cdot \frac{dy}{dx}+\ln y)=e^x\cdot \frac{dy}{dx}+ye^x$

$3x^{2}-\frac{x}y{}\cdot \frac{dy}{dx}-\ln y=e^x\cdot \frac{dy}{dx}+ye^x$

Now we will group the dy/dx terms and move everything else to the opposite side.

$3x^2-\ln y-ye^x=e^x\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}=(e^x+\frac{x}{y})\frac{dy}{dx}$

Then, we can solve for dy/dx.

$\frac{3x^2-\ln y-ye^x}{e^x+\frac{x}{y}} =\frac{dy}{dx}$

To remove the compound fraction, we can multiply the top and bottom of the fraction by y.

$\frac{y(3x^2-\ln y-ye^x)}{y(e^x+\frac{x}{y})} =\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{(3x^2y-y\ln y-y^2e^x)}{e^xy+x}$

The (ugly) answer is $\frac{dy}{dx}=\frac{(3x^2y-y\ln y-y^2e^x)}{e^xy+x}$ .

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Question

Find the second derivative of g(x)

Answer

Find the second derivative of g(x)

To find this derivative, we need to use the product rule:

So, let's begin:

$g'(x)=12x^3ln(x)+3x^4(\frac{1}{x})=12x^3ln(x)+3x^3$

So, we are closer, but we need to derive again to get the 2nd derivative

$g''(x)=36x^2ln(x)+12x^3(\frac{1}{x})+9x^2=36x^2ln(x)+21x^2$

So, our answer is:

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Question

$g \left (x \right )= \cos \left ( \pi x^{2}\right )$

Evaluate $g ' \left ( \frac{1}{2} \right )$ .

Answer

To find , substitute $u = \pi x^{2}$ and use the chain rule:

$g \left (x \right )= \cos \left ( \pi x^{2}\right )$

$g'(x)= \frac{\mathrm{d} }{\mathrm{d} x} \cos \left ( \pi x^{2}\right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} \cos u$

$=\frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \cos u$

$=\frac{\mathrm{d} }{\mathrm{d} x} \pi x^{2} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \cos u$

$= \pi \cdot 2 x\cdot \left ( - \sin u \right )$

$= \pi \cdot 2 x\cdot \left [- \sin \left ( \pi x^{2} ) \right ]$

$=-2 \pi x \sin \left ( \pi x^{2} )$

So $g'(x)=-2 \pi x \sin \left ( \pi x^{2} )$

and

$g ' \left ( \frac{1}{2} \right )=-2 \pi\cdot \frac{1}{2} \sin \left [\pi \cdot \left ( \frac{1}{2} \right )^{2} \right ]$

$=- \pi \sin \left ( \frac{\pi}{4} \right )$

$=- \pi \cdot \frac{\sqrt{2}}{2} = - \frac{ \pi \sqrt{2}}{2}$

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Question

$g (x) = \tan \left ( x +\frac{ \pi }{4}\right )$

Evaluate $g' \left ( \pi )$ .

Answer

To find , substitute $u = x +\frac{ \pi }{4}$ and use the chain rule:

$g (x) = \tan \left ( x +\frac{ \pi }{4}\right )$

$g' (x) = \frac{\mathrm{d} }{\mathrm{d} x}\tan \left ( x +\frac{ \pi }{4}\right )$

$g' (x) = \frac{\mathrm{d} }{\mathrm{d} x}\tan u$

$=\frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \tan u$

$=\frac{\mathrm{d} }{\mathrm{d} x} \left ( x +\frac{ \pi }{4} \right )\cdot \frac{\mathrm{d} }{\mathrm{d} u} \tan u$

$=\frac{\mathrm{d} }{\mathrm{d} x} \left ( x +\frac{ \pi }{4} \right ) \cdot \sec ^{2} u$

$=1 \cdot \sec ^{2}\left ( x +\frac{ \pi }{4}\right ) = \sec ^{2}\left ( x +\frac{ \pi }{4}\right )$

$g'(x) = \sec ^{2}\left ( x +\frac{ \pi }{4}\right )$

$g'(\pi ) = \sec ^{2}\left ( \pi +\frac{ \pi }{4}\right )$

$g'(\pi ) = \sec ^{2}\frac{ 5 \pi }{4}$

$= \left ( - \sqrt{2 }\right )^{2} = 2$

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Question

$f(x)= 0.8x^{3}-0.5x^{2}+ 0.9$

Give .

Answer

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f(x)= 0.8x^{3}-0.5x^{2}+ 0.9$

$f'(x)= 3\cdot 0.8 x^{3-1}-2\cdot 0.5x^{2-1}+ 0$

$f'(x)= 2.4x^{2}-1x^{1}$

$f'(x) = 2.4x^{2}-x$

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Question

$g(x) = e ^{x^{2}+x}$

Evaluate .

Answer

To find , substitute $u = x ^{2} + x$ and use the chain rule:

$g(x) = e ^{x^{2}+x}$

$g'(x) = \frac{\mathrm{d} }{\mathrm{d} x} e ^{x^{2}+x}$

$= \frac{\mathrm{d} }{\mathrm{d} x} e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot e ^{u}$

$= \frac{\mathrm{d} }{\mathrm{d} x}\left ( x ^{2} + x \right ) \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{ x ^{2} + x}$

$g'(x)= \left ( 2x + 1 \right ) e ^{ x ^{2} + x}$

Plug in 3:

$g'(3)= \left ( 2 \cdot 3 + 1 \right ) \cdot e ^{ 3 ^{2} + 3}$

$= \left ( 6 + 1 \right ) \cdot e ^{ 9+ 3} =7e ^{ 12}$

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Question

$g (x) = \frac{1}{4x - 7}$

Evaluate .

Answer

To find , substitute and use the chain rule:

$g (x) = \frac{1}{4x - 7}$

$g '(x) = \frac{\mathrm{d} }{\mathrm{d} x}\cdot \frac{1}{4x - 7}$

$= \frac{\mathrm{d} }{\mathrm{d} x}\cdot \frac{1}{u}$

$= \frac{\mathrm{d} u}{\mathrm{d} x}\cdot \frac{\mathrm{d} }{\mathrm{d} u} \frac{1}{u}$

$= \frac{\mathrm{d} }{\mathrm{d} x} (4x-7) \cdot \frac{\mathrm{d} }{\mathrm{d} u} u ^{-1}$

$=4 \cdot (-1 ) u ^{-1-1}$

$=-4 u ^{-2} = - \frac{4}{u^{2}}= - \frac{4}{(4x-7)^{2}}$

So

$g'(x)= - \frac{4}{(4x-7)^{2}}$

and

$g'(1)= - \frac{4}{(4 \cdot 1 -7)^{2}}$

$= - \frac{4}{(4 -7)^{2}}$

$= - \frac{4}{(-3)^{2}} = - \frac{4}{9}$

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Question

$g (x) = 3 ^{x^{2}}$

Evaluate .

Answer

To find , substitute $u = x ^{2}$ and use the chain rule:

$g (x) = 3 ^{x^{2}}$

$g ' (x) = \frac{\mathrm{d} }{\mathrm{d} x} 3 ^{x^{2}}$

$= \frac{\mathrm{d} }{\mathrm{d} x} 3 ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} 3 ^{u}$

$= \frac{\mathrm{d}}{\mathrm{d} x} x^{2} \cdot \frac{\mathrm{d} }{\mathrm{d} u} 3 ^{u}$

$= 2x \cdot \ln 3 \cdot 3 ^{u}$

$= 2x \ln 3 \cdot 3 ^{x^2}$

So $g'(x)= 2x \ln 3 \cdot 3 ^{x^2}$

and

$g'(-1)= 2(-1) \ln 3 \cdot 3 ^{(-1)^2}$

$= -2 \ln 3 \cdot 3 ^{1} = -2 \ln 3 \cdot 3 = -6 \ln 3$

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Question

Use implicit differentiation to find the slope of the tangent line to $\small y^3-lny=x^2$ at the point $\small (1,1)$ .

Answer

We must take the derivative $\small \left(\small \frac{\mathrm{d} y}{\mathrm{d} x}\right)$ because that will give us the slope. On the left side we'll get

$\small 3y^2\frac{\mathrm{d}y }{\mathrm{d} x}-(1/y)\frac{\mathrm{d} y}{\mathrm{d} x}$ , and on the right side we'll get $\small 2x$ .

We include the $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ on the left side because $\small y$ is a function of $\small x$ , so its derivative is unknown (hence we are trying to solve for it!).

Now we can factor out a $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ on the left side to get

$\small \small \frac{\mathrm{d} y}{\mathrm{d} x}(3y^2-1/y)=2x$ and divide by $\small 3y^2-1/y$ in order to solve for $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ .

Doing this gives you

$\small \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2x}{3y^2-1/y}$ .

We want to find the slope at $\small (1,1)$ , so we can sub in $\small 1$ for $\small x$ and $\small y$ .

$\small \small \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2(1)}{3(1)^2-1/(1)}=2/2=1$ .

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Question

Find dy/dx by implicit differentiation:

$x^3+x\tan^{-1}(y)=e^y$

Answer

To find dy/dx we must take the derivative of the given function implicitly. Notice the term $x\tan^{-1}(y)$ will require the use of the Product Rule, because it is a composition of two separate functions multiplied by each other. Every other term in the given function can be derived in a straight-forward manner, but this term tends to mess with many students. Remember to use the Product Rule:

Product Rule: $\frac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)$

Now if we take the derivative of each component of the given problem statement:

$3x^2\frac{dx}{dx}+[(x)(\frac{1}{1+y^2}(\frac{dy}{dx}))+(1)(\frac{dx}{dx})(\tan^{-1}(y))]=e^y\frac{dy}{dx}$

Notice that anytime we take the derivative of a term with x involved we place a "dx/dx" next to it, but this is equal to "1".

So this now becomes:
$3x^2+[\frac{x}{1+y^2}\frac{dy}{dx}+\tan^{-1}(y)}]=e^y\frac{dy}{dx}$

Now if we place all the terms with a "dy/dx" onto one side and factor out we can solved for it:

$3x^2+\tan^{-1}(y)=e^y\frac{dy}{dx}-\frac{x}{1+y^2}\frac{dy}{dx}$

$3x^2+\tan^{-1}(y)=\frac{dy}{dx}(e^y-\frac{x}{1+y^2})$

$\frac{dy}{dx}=\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

This is one of the answer choices.

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Question

Find dx/dy by implicit differentiation:

$x^3+x\tan^{-1}(y)=e^y$

Answer

To find dx/dy we must take the derivative of the given function implicitly. Notice the term $x\tan^{-1}(y)$ will require the use of the Product Rule, because it is a composition of two separate functions multiplied by each other. Every other term in the given function can be derived in a straight-forward manner, but this term tends to mess with many students. Remember to use the Product Rule:

Product Rule: $\frac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)$

Now if we take the derivative of each component of the given problem statement:

$3x^2\frac{dx}{dy}+[(x)(\frac{1}{1+y^2}(\frac{dy}{dy}))+(1)(\frac{dx}{dy})(\tan^{-1}(y))]=e^y\frac{dy}{dy}$

Notice that anytime we take the derivative of a term with y involved we place a "dy/dy" next to it, but this is equal to "1".

So this now becomes:
$3x^2\frac{dx}{dy}+[\frac{x}{1+y^2}+\tan^{-1}(y)\frac{dx}{dy}]=e^y$

Now if we place all the terms with a "dx/dy" onto one side and factor out we can solved for it:

$3x^2\frac{dx}{dy}+\tan^{-1}(y)\frac{dx}{dy}=e^y-\frac{x}{1+y^2}$

$\frac{dx}{dy}[3x^2+\tan^{-1}(y)]=e^y-\frac{x}{1+y^2}$

$\frac{dx}{dy}=\frac{e^y-\frac{x}{1+y^2}}{3x^2+\tan^{-1}(y)}$

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Question

$\frac{d}{dx} \ln(3x-7)$

Answer

Consider this function a composition of two functions, f(g(x)). In this case, f(x) is ln(x) and g(x) is 3x - 7. The derivative of ln(x) is 1/x, and the derivative of 3x - 7 is 3. The derivative is then $\frac{1}{3x-7} \cdot 3 = \frac{3}{3x-7}$ .

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Question

$\frac{d}{dx } 2 \sin (x^2 )$

Answer

Consider this function a composition of two functions, f(g(x)). In this case, $f(x) =2 \sin (x)$ and . According to the chain rule, $f(g(x)) ' = f'(g(x))\cdot g'(x)$ . Here, $f'(g(x)) = 2 \cos (x^2 )$ and , so the derivative is $2 \cos (x^2 ) \cdot 2x = 4x \cdot \cos(x^2)$

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Question

$\frac{d}{dx } \cos(\ln(x))$

Answer

According to the chain rule, $f(g(x)) ' = f'(g(x))\cdot g'(x)$ . In this case, $f(x) = \cos(x)$ and $g(x) = \ln(x)$ . The derivative is $- \sin (\ln x ) \cdot \frac{1}{x } = \frac{- \sin (\ln x )}{x }$ .

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Question

$\frac{d}{dx } \tan ^ 3 x$

Answer

According to the chain rule, $f(g(x)) ' = f'(g(x))\cdot g'(x)$ . In this case, and $g(x) = \tan x$ . The derivative is $3 (\tan x) ^2 \cdot( 1 + \tan ^2 x ) = 3 \tan ^2 x (1 + \tan ^2 x ) = 3 \tan ^2 x + 3 \tan ^4 x$

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Question

$\frac{d}{dx } 4(x-2)^2$

Answer

According to the chain rule, $f(g(x)) ' = f'(g(x)) \cdot g'(x)$ . In this case, and . and .

The derivative is $8(x-2) \cdot 1 = 8x -16$

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Question

$\frac{d}{dx} e^{\sin x}$

Answer

According to the chain rule, $f(g(x)) ' = f'(g(x)) \cdot g'(x)$ . In this case, and $g(x) = \sin x$ . Since and $g'(x) = \cos x$ , the derivative is $e ^ {\sin x } \cdot \cos x$

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Question

$\frac{\mathrm{d} }{\mathrm{d} x} 3 ^ {x^2 }$

Answer

According to the chain rule, $f(g(x)) ' = f'(g(x)) \cdot g'(x)$ . In this case, and . Since $f'(x) = 3^ {x} \cdot \ln(3)$ and , the derivative is $3^{x^2 } \cdot \ln(3 ) \cdot 2x$

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Question

$\frac{\mathrm{d} }{\mathrm{d} x} \enspace \frac{1}{4} \cdot 2^{4x-1}$

Answer

According to the chain rule, $f(g(x)) ' = f'(g(x)) \cdot g'(x)$ . In this case, $f(x) = \frac{1}{4} \cdot 2^x$ and . Here $f'(x) = \frac{1}{4} \cdot 2^x \cdot \ln(2)$ and . The derivative is:

$\frac{1}{4} \cdot 2 ^{4x-1} \cdot \ln(2) \cdot 4 = 2^{4x-1} \cdot \ln(2)$

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Question

Given the relation , find $\frac{dy}{dx}$ .

Answer

We begin by taking the derivative of both sides of the equation.

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

. (The left hand side uses the Chain Rule.)

.

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

$\frac{dy}{dx} = \frac{2x}{2y-1}$ .

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