The slope of the line tangent to the graph of $\displaystyle f(x) = x^{3} + x - 17$ at the point $\displaystyle (3, 13)$ is $\displaystyle f'(3)$, which can be evaluated as follows:

$\displaystyle f(x) = x^{3} + x - 17$

$\displaystyle f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{3} + x - 17 \right )$

$\displaystyle f'(x) = 3x^{2} + 1$

$\displaystyle f'(3) = 3\cdot 3^{2} + 1 = 28$

The line with slope 28 through $\displaystyle (3, 13)$ has equation:

$\displaystyle y - y_{1} = m (x-x_{1})$

$\displaystyle y -13= 28 (x-3)$

$\displaystyle y -13= 28 x- 28 \cdot 3$

$\displaystyle y -13= 28 x- 84$

$\displaystyle y = 28 x- 71$

This is equivalent to finding a solution of the equation 

$\displaystyle x^9 = 100$, 

or the zero of the polynomial

$\displaystyle P (x) = x^9 - 100$. 

Using Newton's method, we can find $\displaystyle x_{2}$ from the formula

$\displaystyle x_{2} = x_{1} - \frac{P(x_{1})}{P'(x_{1})}$.

$\displaystyle P '(x) = 9 x^8$, so 

$\displaystyle P (x_{1}) = P(2)= 2^9 - 100 = 512 - 100 = 412$

$\displaystyle P '(x_{1}) = P' (2)= 9 \cdot 2^8 = 9 \cdot 256 = 2,304$

$\displaystyle x_{2} = 2 - \frac{P(2)}{P'(2)} = 2 - \frac{412}{2,304} \approx 2 - 0.1788 \approx 1.821$

To find $\displaystyle g'(x)$, substitute $\displaystyle u = x ^{2} + x$ and use the chain rule:

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

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

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

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

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

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

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

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

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

Plug in 3:

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

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

To find $\displaystyle g'(x)$, substitute $\displaystyle u =4x - 7$ and use the chain rule: 

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

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

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

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

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

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

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

So

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

and

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

To find $\displaystyle g'(x)$, substitute $\displaystyle u = x ^{2}$ and use the chain rule: 

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

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

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

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

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

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

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

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

and

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

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

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

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

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

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

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

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

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

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

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

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

and 

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

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

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

The slope of the line tangent to the graph of $\displaystyle f(x)= x^3-2$ at $\displaystyle (2,6)$ is

$\displaystyle f'(2)$, which can be evaluated as follows:

$\displaystyle f(x)= x^3-2$

$\displaystyle f'(x)= 3x^{3-1}-0$

$\displaystyle f'(x)= 3x^{2}$

Then $\displaystyle f'(2)= 3 \cdot 2^{2} = 3 \cdot4 = 12$, which is the slope of the line.

The equation of the line with slope 12 through $\displaystyle (2,6)$ is:

$\displaystyle y - y_{1} = m (x-x_{1})$

$\displaystyle y - 6 = 12 (x-2)$

$\displaystyle y - 6 = 12 x-24$

$\displaystyle y - 6 + 6= 12 x-24 + 6$

$\displaystyle y = 12 x-18$

This is equivalent to finding a solution of the equation 

$\displaystyle x^7 = 1,000$, 

or the zero of the polynomial

$\displaystyle P (x) = x^7 - 1,000$. 

Using Newton's method, we can find $\displaystyle x_{2}$ from the formula

$\displaystyle x_{2} = x_{1} - \frac{P(x_{1})}{P'(x_{1})}$.

$\displaystyle P'(x) = 7x^6$, so

$\displaystyle P (x_{1}) =P(3)= 3^7 - 1,000 = 2,187 - 1,000 = 1,187$

$\displaystyle P'(x_{1}) = P'(3)= 7 \cdot 3^6 = 7 \cdot 729 = 5,103$

$\displaystyle x_{2} = 3 - \frac{P(3)}{P'(3)} = 3 - \frac{1,187}{5,103} \approx 3 - 0.2326 \approx 2.767$

Rewrite the equation to be solved for $\displaystyle x$ as 

$\displaystyle x^3 - x - 19 = 0$.

Let $\displaystyle f(x) = x^3 - x - 19$.  

Then,

$\displaystyle f'(x) = \frac{\mathrm{d} }{\mathrm{d} x} \left ( x^3 - x - 19 \right )$.

$\displaystyle f'(x) =3x^2 - 1$

The problem amounts to finding a zero of $\displaystyle f(x )$. By Newton's method, the second approximation can be derived from the first using the equation

$\displaystyle x_{2} = x_{1} - \frac{f(x_{1})}{f'(x_{1})}$.

$\displaystyle x_{1} = 2.5$, so

$\displaystyle f(2.5) = 2.5^3 - 2.5 - 19 = 15.625 - 2.5 - 19 = -5.875$

and 

$\displaystyle f'(2.5) =3 \cdot 2.5^2 - 1 = 3 \cdot 6.25 - 1 = 18.75 - 1 = 17.75$

$\displaystyle x_{2} = 2.5 - \frac{f(2.5)}{f'(2.5)}= 2.5 - \frac{ -5.875}{17.75}\approx 2.5 - (-0.3310) \approx 2.831$