I need help finding the derivative function of this function

Question

How can I find the derivative of this function ? $$f(x)= (4x^2 + 2x +5)^{0.5}$$

Answer 1

\begin{align} \frac{d}{dx}(4x^2+2x+5)^{0.5} &= 0.5(4x^2+2x+5)^{-0.5}\frac{d}{dx}(4x^2+2x+5) \\ &= 0.5(4x^2+2x+5)^{-0.5}(8x+2) \\ &= \frac{4x+1}{\sqrt{4x^2+2x+5}} \end{align}

Answer 2

This is a composite function so you have to use the chain rule.

The outer function is $x^{0.5}$ and the inner function is $4x^2+2x+5$

The formula is derivative of the outer function evaluated at the inner function multiplied by derivative of the inner function.

Thus your answer is $$ f'(x) = 0.5 (4x^2+2x+5)^{-0.5}(8x+2)$$

You may simplify it to $$ f'(x) = (4x^2+2x+5)^{-0.5}(4x+1)=\frac {4x+1}{\sqrt { 4x^2+2x+5 }} $$