Differentiate, 

$\displaystyle y = \sin(\sin(\sin\(3x)))$

Strategy 

This one at first glance appears difficult even if we recognize that the chain rule is needed; we have a function within a function within a function within a function. To avoid making mistakes, it's best to start by defining variables to make the calculation easier to follow. 

Let's start with the outermost function, we will write $\displaystyle y$ as a function of $\displaystyle u$ by setting, 

 $\displaystyle u = \sin(\sin(3x))$

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$\displaystyle y = \sin(\underset{u}{ \underbrace{\sin(\sin(3x))}}) = \sin(u)$

$\displaystyle y = \sin(u)$

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Similarly, define $\displaystyle v$ to write $\displaystyle u$ as a function of $\displaystyle v$ 

 $\displaystyle u = \sin(\underset{v}{ \underbrace{\sin(3x)}}) = \sin(v)$ 

$\displaystyle u = \sin(v)$

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Write $\displaystyle v$ as a function of $\displaystyle w$

$\displaystyle v = \sin(\underset{w}{ \underbrace{3x}} )=\sin(w)$

$\displaystyle v = \sin(w)$

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Finally, define the inner-most function, $\displaystyle w$, as the function of $\displaystyle x$

$\displaystyle w=3x$

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 $\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dv}\frac{dv}{dw}\frac{dw}{dx}$

Since $\displaystyle \frac{dw}{dx}=3$ we will just substitute that in and move to the front. 

$\displaystyle \frac{dy}{dx}=3\cos(u)\cos(v)\cos(w)$

That was easy enough, now just write everything in terms of $\displaystyle x$ by going back to the definitions of $\displaystyle u,$ $\displaystyle v,$ and $\displaystyle w$.

$\displaystyle u = \sin(\sin(3x))$

$\displaystyle v = \sin(3x)$

$\displaystyle w = 3x$

 $\displaystyle \frac{dy}{dx}=3\cos(\sin(\sin(3x)))\times \cos(\sin(3x))\times \cos(3x)$

To find the tangent line at the given point, we need to first take the derivative of the given function. 

Power Rule:

The Power Rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent.

Therefore, $\displaystyle y= x^2-3x+4$ becomes $\displaystyle y' = 2x - 3$

From there we plug in the "1" from the point to get our m value of the equation $\displaystyle y=mx + b$. When we plug in "1" to y' we get m=-1. Then from there, we will plug our point into $\displaystyle y=mx + b$ now that we have found m to find our b value. So,

$\displaystyle 2=(-1)(1) + b$

$\displaystyle b= 3$

Therefore, the tangent line is equal to

$\displaystyle y= -x+3$

To find the tangent line at the given point, we need to first take the derivative of the given function. The rule for functions with "e" in it says that the derivative of $\displaystyle e^x=e^x$ However with this function there is also a 3 in the exponent so we will also use chain rule. Chain Rules states that we work from the outside to the inside. Meaning we will take the derivative of the outside of the equation and multiply it by the derivative of the inside of the equation.

To put this into equation it will look like $\displaystyle 3e^{3x}$

From there we plug in the "0" from the point to get our m value of the equation . When we plug in "0" to y' we get m=3. Then from there, we will plug our point into $\displaystyle y=mx + b$  now that we have found m to find our m value. So,

$\displaystyle 1=3(0)+b$

$\displaystyle b=1$ then plug this all back into the equation once more and we are left with

$\displaystyle y=3x + 1$

To find the tangent line at the given point, we need to first take the derivative of the given function using Power Rule

Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent. $\displaystyle (nx^{^{n-1}})$

From there we plug in the "2" from the point to get our m value of the equation. When we plug in "2" to y' we get m=8. Then from there, we will plug our point into now that we have found m to find our b value. So,

$\displaystyle y=mx+b$

$\displaystyle 4=8(2) + b ; b=-12$

Plug this back into $\displaystyle y=mx+b$

$\displaystyle y= 8x - 12$

Find the equation of the line tangent to the curve $\displaystyle f$ at the given point

$\displaystyle f(x)=\tan(x)+x$           $\displaystyle x = \frac{\pi}{4}$

The slope of the line tangent at the given point will be equal to the derivative of $\displaystyle f(x)$ at that point. Compute the derivative and find the slope for our line: 

$\displaystyle f'(x)=\sec^2(x)+1$

$\displaystyle f'\left(\frac{\pi}{4} \right )=\sec^2\left(\frac{\pi}{4}\right)+1$

Evaluate the secant term: 

$\displaystyle \sec^2\left(\frac{\pi}{4}\right)=\frac{1}{\cos^2\left(\frac{\pi}{4}\right)}=\left(\frac{2}{\sqrt{2}} \right )^2=2$ 

Therefore slope of the tangent line is simply: 

$\displaystyle f'\left(\frac{\pi}{4} \right )=3$

So now we know the slope of the tangent line and can write the equation then solve for $\displaystyle b.$ 

$\displaystyle y = 3x+b$

In order to solve for $\displaystyle b$ we need one point on the line. Use the point where the tangent line meets the curve. Use the original function to find the "y" coordinate at this point: 

$\displaystyle f\left(\frac{\pi}{4} \right )\: \: =\: \: \tan\left(\frac{\pi}{4}\right) +\frac{\pi}{4}\: \:=\: \: 1+\frac{\pi}{4}\: \: =\: \: \frac{4+\pi}{4}$

We now have our point:

 $\displaystyle \left(\frac{\pi}{4}\: \: ,\: \frac{4+\pi}{4}\right )$

Use the point to find $\displaystyle b$

$\displaystyle \frac{4+\pi}{4}=3\left(\frac{\pi}{4} \right )+b$

$\displaystyle b = \frac{2-\pi}{2}$

$\displaystyle y = 3x+\frac{2-\pi}{2}$

Find the slope of the line tangent to the curve of d(g) when g=6.

$\displaystyle d(g)=14g^3+6g^2-2g+12$

All we need here is the power rule. This states that to find the derivative of a polynomial, simply subtract 1 from each exponent and then multiply each term by their original exponent. 

Constant terms will drop out when we do this, and linear terms will become constants.

$\displaystyle d(g)=14g^3+6g^2-2g+12$

$\displaystyle d'(g)=(3)14g^2+(2)6g-2=42g^2+12g-2$

$\displaystyle d'(g)=42g^2+12g-2$

From here substitute in g=6.

$\displaystyle \\d'(6)=42(6)^2+12(6)-2 \\d'(6)=1582$

The tangent line to the graph of $\displaystyle f$ at point $\displaystyle (2, 9)$ is the line with slope $\displaystyle f'(2)$ that passes through that point. Find the derivative $\displaystyle f'(x)$:

$\displaystyle f(x) = 4x+ e^{x-2}$

$\displaystyle f(x) = 4x+ \frac{1}{e^{2}} \cdot e^{x}$

$\displaystyle f'(x) =\frac{\mathrm{d} }{\mathrm{d} x} \left (4x+ \frac{1}{e^{2}} \cdot e^{x} \right )$

Apply the sum rule:

$\displaystyle f'(x) =\frac{\mathrm{d} }{\mathrm{d} x} (4x ) + \frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{1}{e^{2}} \cdot e^{x} \right )$

$\displaystyle f'(x) =\frac{\mathrm{d} }{\mathrm{d} x} (4x ) + \frac{1}{e^{2}} \frac{\mathrm{d} }{\mathrm{d} x} \left ( e^{x} \right )$

$\displaystyle f'(x) =4 + \frac{1}{e^{2}} e^{x}$

$\displaystyle f'(x) =4 + e^{x-2}$

$\displaystyle f'(2)= 4 + e^{2-2} = 4 + e^{0} = 4 + 1 = 5$

The tangent line is therefore the line with slope 5 through $\displaystyle (2, 9)$. Apply the point-slope formula:

$\displaystyle y - 9 = 5(x-2)$

$\displaystyle y - 9 =5x-10$

$\displaystyle y = 5x-1$

The tangent line to the graph of $\displaystyle f$ at point $\displaystyle (0, 2)$ is the line with slope $\displaystyle f'(0)$ that passes through that point. Find the derivative $\displaystyle f'(x)$:

$\displaystyle f(x) =3 \sin \frac{x}{2} +2 \cos \frac{x}{3}$

$\displaystyle f'(x) =\frac{\mathrm{d} }{\mathrm{d} x}\left (3 \sin \frac{x}{2} +2 \cos \frac{x}{3} \right )$

Apply the constant multiple and sum rules:

$\displaystyle f'(x) =\frac{\mathrm{d} }{\mathrm{d} x}\left (3 \sin \frac{x}{2} +2 \cos \frac{x}{3} \right )$

$\displaystyle f'(x) =\frac{\mathrm{d} }{\mathrm{d} x}\left (3 \sin \frac{x}{2} \right )+ \frac{\mathrm{d} }{\mathrm{d} x}\left ( 2 \cos \frac{x}{3} \right )$

$\displaystyle f'(x) =3 \frac{\mathrm{d} }{\mathrm{d} x}\left ( \sin \frac{x}{2} \right )+2 \frac{\mathrm{d} }{\mathrm{d} x}\left ( \cos \frac{x}{3} \right )$

Set $\displaystyle u = \frac{x}{2}$ and $\displaystyle v= \frac{x}{3}$ and apply the chain rule. 

$\displaystyle f'(x) =3 \frac{\mathrm{d} }{\mathrm{d} x}\left ( \sin u \right )+2 \frac{\mathrm{d} }{\mathrm{d} x}\left ( \cos v \right )$

$\displaystyle f'(x) =3 \frac{\mathrm{d} u}{\mathrm{d} x}\cdot \frac{\mathrm{d} }{\mathrm{d} u} \left ( \sin u \right )+2 \frac{\mathrm{d} v}{\mathrm{d} x}\cdot \frac{\mathrm{d} }{\mathrm{d} v} \left ( \cos v \right )$

$\displaystyle f'(x) =3 \frac{\mathrm{d} u}{\mathrm{d} x}\cdot \cos u +2 \frac{\mathrm{d} v}{\mathrm{d} x}\cdot (- \sin v)$

Substituting back:

$\displaystyle f'(x) =3 \frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{x}{2} \right )\cdot \cos \frac{x}{2} -2 \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{x}{3} \right )\cdot \sin \frac{x}{3}$

$\displaystyle f'(x) =3 \left ( \frac{1}{2} \right )\cdot \cos \frac{x}{2} -2 \left ( \frac{1}{3} \right )\cdot \sin \frac{x}{3}$

$\displaystyle f'(x) = \frac{3}{2} \cos \frac{x}{2} - \frac{2}{3} \sin \frac{x}{3}$

Evaluate $\displaystyle f'(0)$ using substitution:

$\displaystyle f'(0) = \frac{3}{2} \cos \frac{0}{2} - \frac{2}{3} \sin \frac{0}{3}$

$\displaystyle f'(0) = \frac{3}{2} \cos 0 - \frac{2}{3} \sin 0$

$\displaystyle f'(0) = \frac{3}{2}(1) - \frac{2}{3}( 0)$

$\displaystyle f'(0) = \frac{3}{2}$

The tangent line is therefore the line with slope $\displaystyle \frac{3}{2}$ through $\displaystyle (0, 2)$. $\displaystyle (0, 2)$ is a $\displaystyle y$-intercept, so apply the slope-intercept formula to get the equation

$\displaystyle y = \frac{3}{2}x+2$.

This is not among the choices given.

We first start by finding the slope of the line in question, which we do by taking the derivative of $\displaystyle y=4x^3+2$ and evaluate at $\displaystyle x=1$. 

$\displaystyle y'=12x^2$, $\displaystyle y'(1)=12=m$

We then use point slope form to get the equation of the line at the point $\displaystyle (2,4)$

$\displaystyle (y-4)=12(x-2)$

$\displaystyle y=12x-20$

The first step is to find the derivative of the function given, which is $\displaystyle f'(x)=3x^{2}-6x+2$. Next, find the slope at (1,4) by plugging in x=1 and solving for $\displaystyle f'(x)$, which is the slope. You should get $\displaystyle f'(1)= -1$. This means the slope of the new line is also -1 because at the point where a slope and a line are tangent they have the same slope. Use the equation $\displaystyle y=mx+b$ to express your line. Y and x are variables and m is the slope, so the only thing you need to find is b. Plug in the point and slope into $\displaystyle y=mx+b$to get $\displaystyle b=5$. Now you can express the general equation of the line as $\displaystyle y=-x+5$.