# What's the derivative of f(x) = 2arcsin(e^(3x))?

Oct 22, 2016

$f ' \left(x\right) = \frac{6 {e}^{3 x}}{\sqrt{1 - \textcolor{b r o w n}{{e}^{6 x}}}}$

#### Explanation:

$f \left(x\right)$ is a composite of two functions $\textcolor{b l u e}{g \left(x\right) = \arcsin x}$ and $\textcolor{b r o w n}{h \left(x\right) = {e}^{3 x}}$

We have: $f \left(x\right) = 2 g \circ h \left(x\right)$
Then chain rule is applied: color(red)(f'(x)=2*g'(h(x))*h'(x)

Let us compute $\textcolor{red}{g ' \left(h \left(x\right)\right)}$
$\textcolor{b l u e}{g \left(x\right) = \arcsin x}$
color(blue)(g'(x))=1/(sqrt(1-x^2)
Then,
color(red)(g'(h(x)))=1/(sqrt(1-(h(x))^2)
color(red)(g'(h(x)))=1/(sqrt(1-color(brown)((e^(3x))^2)
color(red)(g'(h(x)))=1/(sqrt(1-color(brown)(e^(6x)

Let us compute $\textcolor{red}{h ' \left(x\right)}$
$\textcolor{b r o w n}{h \left(x\right) = {e}^{3 x}}$
Knowing that the derivative of exponential function is as follows:
$\left({e}^{u \left(x\right)}\right) ' = u ' \left(x\right) {e}^{u \left(x\right)}$

$\textcolor{red}{h ' \left(x\right)} = \left({e}^{3 x}\right) '$
$\textcolor{red}{h ' \left(x\right)} = 3 {e}^{3 x}$

color(red)(f'(x)=2*g'(h(x))*h'(x)
$\textcolor{red}{f ' \left(x\right)} = 2 \cdot \frac{1}{\sqrt{1 - \textcolor{b r o w n}{{e}^{6 x}}}} \cdot 3 {e}^{3 x}$
$\textcolor{red}{f ' \left(x\right)} = \frac{6 {e}^{3 x}}{\sqrt{1 - \textcolor{b r o w n}{{e}^{6 x}}}}$