Derivative of $e^\sqrt{4x+4}$

Question

$$f(x)=e^\sqrt{4x+4}$$

$f(x)=e^u$
$u=\sqrt{4x+4}=(4x+4)^{1/2}$
$u\;'=\dfrac{1}{2}(4x+4)^{-1/2}=\dfrac{1}{2\sqrt{4x+4}}$

I don't know how to proceed from here. Thanks.

Answer 1

Just write the base and chain along the derivative of the exponent: \begin{equation} \frac{d}{dx}\left(e^{\sqrt{4x+4}}\right)=e^{\sqrt{4x+4}}\cdot\frac{1}{2}\left(4x+4\right)^{\frac{-1}{2}}\left(4\right)=\boxed{\frac{2e^{\sqrt{4x+4}}}{\sqrt{4x+4}}.} \end{equation}

Answer 2

$$f'(x)=e^\sqrt{4x+4}\times \frac12 \frac{1}{\sqrt{4x+4}} \times 4=e^\sqrt{4x+4}\times \frac{1}{\sqrt{x+1}}.$$

Notice that $$(f(g(x)))'=f'(g(x))\times g'(x)$$

Answer 3

Hint : $\dfrac {\operatorname d \!e^{f(x)}}{\operatorname d \!x}=e^{f(x)} \dfrac {\operatorname d \!f}{\operatorname d \!x}$