# Find the derivative of the function g(x) = \arctan x + \frac{x}{\sqrt{1+x^2}}

## Question:

Find the derivative of the function

{eq}g(x) = \arctan x + \frac{x}{\sqrt{1+x^2}} {/eq}

## Quotient Rule:

The quotient rule is a useful rule for derivatives that we can apply whenever our function is a ratio of other functions, i.e. of the form {eq}h (x) = \frac{f (x)}{g (x)} {/eq}. Then the derivative is given by

{eq}\displaystyle \left[ \frac fg \right]' = \frac{f'g - fg'}{g^2} {/eq}

Note that this is really just a generalized form of the product rule where one the functions has a negative exponent.

Recall that

{eq}\begin{align*} \frac{d}{dx} \left( \arctan x \right) &= \frac1{1+x^2} \end{align*} {/eq}

Then we have

{eq}\begin{align*} g' (x) &= \frac{d}{dx} \left( \arctan x + \frac{x}{\sqrt{1+x^2}} \right) \\ &= \frac1{1+x^2} + \frac{\frac{d}{dx}(x) \cdot (1+x^2)^{1/2} - x \cdot \frac12(1+x^2)^{-1/2}}{1+x^2} \\ &= \frac1{1+x^2} + \frac{1 \cdot (1+x^2)^{1/2} - x \cdot \frac12(1+x^2)^{-1/2}(2x)}{1+x^2} \\ &= \frac1{1+x^2} + \frac{(1+x^2) - x^2 }{(1+x^2)^{3/2}} \\ &= \frac{\sqrt{1+x^2}+1}{(1+x^2)^{3/2}} \\ \end{align*} {/eq}