Answer
$$\frac{e^{x} \sin ^{-1} x}{\ln x}\left( 1+\frac{1}{\sin^{-1}x\sqrt{1-x^2}} -\frac{1 }{x\ln x} \right)$$
Work Step by Step
Given
$$y=\frac{e^{x} \sin ^{-1} x}{\ln x}$$ Since \begin{align*} \ln y&=\ln \frac{e^{x} \sin ^{-1} x}{\ln x}\\ &=\ln( e^x\sin^{-1}x )-\ln \ln x\\ &= \ln( e^x)+\ln (\sin^{-1}x )-\ln \ln x \end{align*} Differentiate both sides \begin{align*} \frac{1}{y}y'&= \frac{1}{e^x}e^x+\frac{1}{\sin^{-1}x}\frac{1}{\sqrt{1-x^2}} -\frac{1}{\ln x} \frac{1}{ x}\\ &= 1+\frac{1}{\sin^{-1}x\sqrt{1-x^2}} -\frac{1 }{x\ln x} \\ y'&=\frac{e^{x} \sin ^{-1} x}{\ln x}\left( 1+\frac{1}{\sin^{-1}x\sqrt{1-x^2}} -\frac{1 }{x\ln x} \right) \end{align*}
