Differentiating this inverse trigonometric function $$\sin^{-1}\left( \frac{2^{x+1}\cdot3^x}{1+36^{x}} \right)$$
Had this question for todays test but still cannot find out how to proceed.

The function
$$
f(x)=\arcsin\frac{2x}{1+x^2}
$$
is defined for every real $x$, because $|2x|<1+x^2$. We also have
$$
f'(x)=\frac{1}{\sqrt{1-\left(\dfrac{2x}{1+x^2}\right)^2}}
\cdot 2\frac{(1+x^2)-x\cdot 2x}{(1+x^2)^2}=
2\frac{1+x^2}{|1-x^2|}\frac{1-x^2}{(1+x^2)^2}
$$
so
$$
f'(x)=\begin{cases}
\dfrac{2}{1+x^2} & \text{if $|x|<1$} \\[6px]
-\dfrac{2}{1+x^2} & \text{if $|x|>1$}
\end{cases}
$$
Thus we can write
$$
f(x)=\begin{cases}
a+2\arctan x & \text{if $|x|<1$} \\[6px]
b-2\arctan x & \text{if $|x|>1$}
\end{cases}
$$
Since $f(0)=0$, we have $a=0$. Since $\lim_{x\to\infty}f(x)=0$ we have $b=\pi$. Finally
$$
f(x)=\begin{cases}
2\arctan x & \text{if $|x|\le 1$} \\[6px]
\pi-2\arctan x & \text{if $|x|>1$}
\end{cases}
$$
The function is not differentiable at $-1$ and $1$.
You can now write your complicated function as
$$
g(x)=\arcsin\frac{2^{x+1}\cdot 3^x}{1+36^x}=f(6^x)
$$
by noticing that $2^{x+1}\cdot 3^x=2\cdot 6^x$ and $36^x=(6^x)^2$.
Thus, by the chain rule,
$$
g'(x)=f'(6^x)\cdot 6^x\ln 6
$$
provided $6^x\ne1$, that is, $x\ne0$. Explicitly,
$$
g'(x)=\begin{cases}
\dfrac{2\cdot 6^x\cdot \ln 6}{1+36^x} & \text{if $x>0$} \\[6px]
-\dfrac{2\cdot 6^x\cdot \ln 6}{1+36^x} & \text{if $x<0$} \\[6px]
\end{cases}
$$
The function is not differentiable at $0$, as it can be seen in the diagram (and proved directly).

