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.

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 $$ y' = \ln 6 \tan y \left(\frac{1 - 36^x}{1 + 36^x}\right)$$ 
