Computation of Derivatives - AP Calculus BC

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Question

$\newline {\vec{r}(t)} = (3e^t+t^3,e^{2t}+t,9e^t+e^{2t})\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

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Answer

In general:

If $\newline \ {\vec{r}(t)} = (f(t),g(t),z(t))$ ,

then $\newline \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (\frac{\mathrm{d} }{\mathrm{d} t}f(t), \frac{\mathrm{d} }{\mathrm{d} t}g(t),\frac{\mathrm{d} }{\mathrm{d} t}z(t))$

Derivative rules that will be needed here:

Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$

When taking derivatives of sums, evaluate with the sum rule which states that the derivative of the sum is the same as the sum of the derivative of each term: $\frac{\mathrm{d} }{\mathrm{d} t}(t^n+10) = \frac{\mathrm{d} }{\mathrm{d} t}t^n + \frac{\mathrm{d} }{\mathrm{d} t}10 = n*t^{n-1}$

Special rule when differentiating an exponential function: $\frac{\mathrm{d} }{\mathrm{d} t} e^{kt} = ke^{kt}$ , where k is a constant.

In this problem, ${\vec{r}(t)} = ((3e^t + t^3),(e^{2t} + t),(9e^t+e^{2t}))$

$\frac{\mathrm{d} }{\mathrm{d} t}(3e^t + t^3)= \frac{\mathrm{d} }{\mathrm{d} t}3e^t + \frac{\mathrm{d} }{\mathrm{d} t}t^3 = 3e^t + 3t^2$

$\frac{\mathrm{d} }{\mathrm{d} t}(e^{2t} + t)= \frac{\mathrm{d} }{\mathrm{d} t}e^{2t}+ \frac{\mathrm{d} }{\mathrm{d} t}t =2e^{2t} + 1$

$\frac{\mathrm{d} }{\mathrm{d} t}(9e^t+e^{2t})= \frac{\mathrm{d} }{\mathrm{d} t}9e^t + \frac{\mathrm{d} }{\mathrm{d} t}e^{2t} = 9e^t + 2e^{2t}$

Put it all together to get $\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3e^t + 3t^2,2e^{2t} + 1,9e^t + 2e^{2t})$

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Question

Find the derivative of the following set of parametric equations:

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Answer

We start by taking the derivative of x and y with respect to t, as both of the equations are only in terms of this variable:

$\frac{dx}{dt}=6t^2+10t-11$

$\frac{dy}{dt}=14t-4$

The problem asks us to find the derivative of the parametric equations, dy/dx, and we can see from the work below that the dt term is cancelled when we divide dy/dt by dx/dt, leaving us with dy/dx:

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

So now that we know dx/dt and dy/dt, all we must do to find the derivative of our parametric equations is divide dy/dt by dx/dt:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{14t-4}{6t^2+10t-11}=\frac{2(7t-2)}{6t^2+10t-11}$

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Question

Find dy/dx by implicit differentiation:

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

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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 the derivative of the following polar equation:

$r=3+8sin\theta$

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Answer

Our first step in finding the derivative dy/dx of the polar equation is to find the derivative of r with respect to $\theta$ . This gives us:

$\frac{dr}{d\theta}=8cos\theta$

Now that we know dr/d $\theta$ , we can plug this value into the equation for the derivative of an expression in polar form:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}sin\theta+rcos\theta }{\frac{dr}{d\theta }cos\theta-rsin\theta }=\frac{8cos\theta sin\theta+(3+8sin\theta )cos\theta}{8cos^2\theta-(3+8sin\theta)sin\theta }$

Simplifying the equation, we get our final answer for the derivative of r:

$\frac{dy}{dx}=\frac{16cos\theta sin\theta +3cos\theta }{8cos^2\theta -3sin\theta -8sin^2\theta }$

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Question

Find dx/dy by implicit differentiation:

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

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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.

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Question

Given that $f: \mathbb{R}^{n}}\rightarrow \mathbb{R}$ . We define its gradient as :

$\nabla f =\left[\frac{\partial f}{\partial x_{1}} \cdots \frac{\partial f}{\partial x_{n}}\right]$

Let $f: \mathbb{R}^{n}}\rightarrow \mathbb{R}$ be given by:

$f(x_{1},x_{2},\cdots, x_{n}})=\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

What is the gradient of ?

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Answer

By definition, to find the gradient vector , we will have to find the gradient components. We know that the gradient components are that partial derivatives.

We know that in our case we have :

$\frac{\partial f}{\partial x_{i}}=\frac{x_{i}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

To see this, fix all other variables and assume that you have only $x_{i}$ as the only variable.

Now we apply the given defintion , i.e,

$\nabla f =\left[\frac{\partial f}{\partial x_{1}} \cdots \frac{\partial f}{\partial x_{n}}\right]$

with :

$\frac{\partial f}{\partial x_{1}}=\frac{x_{1}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

$\frac{\partial f}{\partial x_{2}}=\frac{x_{2}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

$\frac{\partial f}{\partial x_{3}}=\frac{x_{3}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

$\vdots$

$\frac{\partial f}{\partial x_{n}}=\frac{x_{n}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

this gives us the solution .

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Question

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

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

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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 $g' \left ( \pi )$ .

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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

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

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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

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

Evaluate .

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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 .

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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 .

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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 $\small (1,1)$ .

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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

Let $f:\mathbb{R}^{n}\mapsto \mathbb{R}$ .

We define the gradient of as:

$\nabla f =\left[\frac{\partial f}{\partial x_{1}} \cdots \frac{\partial f}{\partial x_{n}}\right]$

Let $f=sin(x_{1}+2x_{2}+\cdots nx_{n})$ .

Find the vector gradient.

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Answer

We note first that :

$\frac{\partial f}{\partial x_{1}}=cos(x_{1}+2x_{2}+\cdots nx_{n})$

Using the Chain Rule where $x_{1}$ is the only variable here.

$\frac{\partial f}{\partial x_{2}}=2 cos(x_{1}+2x_{2}+\cdots nx_{n})$

Using the Chain Rule where $x_{2}$ is the only variable here.

Continuing in this fashion we have:

$\frac{\partial f}{\partial x_{n}}=ncos(x_{1}+2x_{2}+\cdots nx_{n})$

Again using the Chain Rule and assuming that $x_{n}$ is the variable and all the others are constant.

Now applying the given definition of the gradient we have the required result.

$[cos(x_{1}+2x_{2}+\cdots nx_{n}) \quad 2cos(x_{1}+2x_{2}+\cdots nx_{n})\quad \cdots\quad ncos(x_{1}+2x_{2}+\cdots nx_{n})\]$

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Question

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

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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

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

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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

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

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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

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

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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

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

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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

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

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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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