Question

$\frac{e^x+\sin x}{1+\log x}$

Answer 1

$$\begin{gathered} \mathrm{Let}u=e^x+\sin x;v=1+\log x \\ Then,u\text{'}=e^x+\cos x;v\text{'}=\frac{1}{x} \\ \text{Using the quotient rule:} \\ \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu\text{'}-uv\text{'}}{v^2} \\ \frac{d}{dx}\left(\frac{e^x+\sin x}{1+\log x}\right)=\frac{\left(1+\log x\right)\left(e^x+\cos x\right)-\left(e^x+\sin x\right)\left({\displaystyle \frac{1}{x}}\right)}{{\left(1+\log x\right)}^2} \\ =\frac{x\left(1+\log x\right)\left(e^x+\cos x\right)-\left(e^x+\sin x\right)}{x{\left(1+\log x\right)}^2} \end{gathered}$$