# a. Find the derivative of the function. f\left ( x \right ) = cos\left ( \arctan \ 6x \right ) ...

## Question:

a. Find the derivative of the function.

{eq}f\left ( x \right ) = cos\left ( \arctan \ 6x \right ) {/eq}

b. Find the equation of the tangent line to the graph of {eq}f\left ( x \right ) = \frac{4x - 7}{x + 1} {/eq} at the point at which x = 0.

## Rules of Differentiation:

The quotient rule states: {eq}\left( \dfrac f g \right)' = \dfrac{f'g - fg'}{g^2} {/eq}

The chain rule states: {eq}\left( f(g(x) \right)' = f'(g(x)) \cdot g'(x) {/eq}

Let's use these rules to evaluate the following derivative.

We will also use the following facts:

{eq}\displaystyle \dfrac{d}{dx} \cos x = -\sin x \\ \displaystyle \dfrac{d}{dx} \arctan x = \dfrac 1 {1+x^2} {/eq}

a. {eq}f\left ( x \right ) = cos\left ( \arctan \ 6x \right ) {/eq}

We use the chain rule twice:

{eq}\begin{align*} f'(x) & = -\sin(\arctan 6x) \cdot(\arctan (6x))' \\&= -\sin(\arctan 6x) \cdot \dfrac{1}{1+(6x)^2} \cdot (6x)' \\& = -\sin(\arctan 6x) \cdot \dfrac{1}{1+(6x)^2} \cdot 6 \\& = -6\sin(\arctan 6x) \cdot \dfrac{1}{1+(6x)^2} \end{align*} {/eq}

b. {eq}f\left ( x \right ) = \frac{4x - 7}{x + 1} {/eq}

We use the quotient rule to differentiate:

{eq}\begin{align*} f'(x) & =\dfrac{(4x-7)'(x+1) - (4x-7)(x+1)'} {(x+1)^2} \\& = \dfrac{4(x+1) - (4x-7)(1) }{(x+1)^2} \\& = \dfrac{4x+4-4x+7}{(x+1)^2} \\& = \dfrac{11}{(x+1)^2} \end{align*} {/eq}

Hence the slope of the tangent line at x=0 is:

{eq}f'(0) = \dfrac {11}{(0+1)^2} = 11 {/eq}

Furthermore, when x=0,

{eq}f(0) = \frac{4\cdot 0 - 7}{0 + 1} = -7 {/eq}

Hence the equation of the tangent line at x=0 is:

{eq}y - (-7) = 11(x-0) \\ y +7 = 11x \\ y = 11x-7 {/eq}