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.
 A: 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).

A: Let
\begin{align}
y&=\sin ^{-1}\left(\frac{2^{x+1}3^x}{1+36^x}\right)\\
 &=\sin ^{-1}\left(\frac{2^x\cdot2\cdot3^x}{1+36^x}\right)\\
 &=\sin ^{-1}\left(\frac{2\cdot6^x}{1+6^{2x}}\right)\\
&=2\tan ^{-1}6^x
\end{align}
Thus
\begin{align}
\frac{dy}{dx}&=\frac{2}{1+6^{2x}}\cdot6^x\log 6\\
\implies \frac{dy}{dx}&=\frac{2\cdot6^x\log 6}{1+36^x}
\end{align}
A: If we have $y = \arcsin(2^{x + 1} 3^x / (1 + 36^x))$, then $$\sin y = \frac{2^{x + 1} 3^x}{1 + 36^x}$$ The expression on the right is still a daunting product/quotient, but it can be reduced to sums/differences using logarithms, giving $$\ln \sin y = (x + 1) \ln 2 + x \ln 3 - \ln(1 + 36^x)$$ Now it is routine to take the derivative with respect to $x$ (remember the chain rule for the left!): $$ \frac{y' \cos y}{\sin y} = \ln 2 + \ln 3 - \frac{36^x \ln(36)}{1 + 36^x} = \ln 6 \left(1 - \frac{2 \cdot 36^x}{1 + 36^x}\right) $$ and solving for $y'$ gives $$ y' = \ln 6 \tan y \left(\frac{1 - 36^x}{1 + 36^x}\right)$$ 
A: you need the caine rule and the rule for an exponential function
$$\frac{\frac{2^{x+1} 3^x \log (3)}{36^x+1}+\frac{2^{x+1} 3^x \log (2)}{36^x+1}-\frac{2^{3
   x+1} 27^x \log (36)}{\left(36^x+1\right)^2}}{\sqrt{1-\frac{2^{2 x+2} 3^{2
   x}}{\left(36^x+1\right)^2}}}$$
A: I'll explain a few steps you can use to solve your problem:


*

*$$\frac{\text{d}\arcsin(f(x))}{\text{d}x}=\frac{f'(x)}{\sqrt{1-f(x)^2}}$$

*When $f(x)=\frac{z(x)}{y(x)}$ we get:
$$f'(x)=\frac{\text{d}\left(\frac{z(x)}{y(x)}\right)}{\text{d}x}=\frac{y(x)z'(x)-z(x)y'(x)}{y(x)^2}$$

*When $y(x)=\text{k}+\text{c}^{q(x)}$ we get ($\text{c}$ and $\text{k}$ are constants):
$$y'(x)=\frac{\text{d}\left(\text{k}+\text{c}^{q(x)}\right)}{\text{d}x}=q'(x)\ln(\text{c})\text{c}^{q(x)}$$

*When $z(x)=r(x)\cdot v(x)$ we get:
$$z'(x)=\frac{\text{d}\left(r(x)\cdot v(x)\right)}{\text{d}x}=v(x)r'(x)+r(x)v'(x)$$


Now, you can use:


*

*$$\frac{\text{d}\left(x\right)}{\text{d}x}=1$$

*When $\text{a}$ is a constant:
$$\frac{\text{d}\left(\text{a}\right)}{\text{d}x}=0$$

