Computation of Derivatives - AP Calculus BC

Card 1 of 790

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

Find the second derivative of g(x)

Tap to reveal answer

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:

← Didn't Know|Knew It →

Question

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

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

Tap to reveal answer

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

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

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

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

← Didn't Know|Knew It →

Question

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

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

Tap to reveal answer

Answer

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

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

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

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

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

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

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

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

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

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

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

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

← Didn't Know|Knew It →

Question

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

Give .

Tap to reveal answer

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

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

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

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

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

← Didn't Know|Knew It →

Question

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

Evaluate .

Tap to reveal answer

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:

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

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

← Didn't Know|Knew It →

Question

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

Evaluate .

Tap to reveal answer

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

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

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

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

← Didn't Know|Knew It →

Question

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

Evaluate .

Tap to reveal answer

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

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

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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 $\small y$ .

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

← Didn't Know|Knew It →

Question

Find dy/dx by implicit differentiation:

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

Find dx/dy by implicit differentiation:

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

Tap to reveal answer

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:

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

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

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

This is one of the answer choices.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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 $\frac{1}{3x-7} \cdot 3 = \frac{3}{3x-7}$ .

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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 $2 \cos (x^2 ) \cdot 2x = 4x \cdot \cos(x^2)$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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 $- \sin (\ln x ) \cdot \frac{1}{x } = \frac{- \sin (\ln x )}{x }$ .

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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 $3 (\tan x) ^2 \cdot( 1 + \tan ^2 x ) = 3 \tan ^2 x (1 + \tan ^2 x ) = 3 \tan ^2 x + 3 \tan ^4 x$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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 $e ^ {\sin x } \cdot \cos x$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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 $3^{x^2 } \cdot \ln(3 ) \cdot 2x$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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:

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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.)

.

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

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

← Didn't Know|Knew It →

Question

Given the relation , find .

Tap to reveal answer

Answer

We can use implicit differentiation to find . We being by taking the derivative of both sides of the equation.

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

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

$2x +[x \times 2y' + y^2 \times 1] = y'$ (This line uses the product rule for the derivative of .)

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

← Didn't Know|Knew It →

0

Knew It