Question

$\mathrm{If}y=x^{\tan x}+\sqrt{\frac{x^2+1}{2}},\mathrm{find}\frac{dy}{dx}$

Answer 1

$$\begin{gathered} y=x^{\tan x}+\sqrt{\frac{x^2+1}{2}} \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \log y=\tan x\log x+\frac{1}{2}\log \left(\frac{x^2+1}{2}\right) \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{\tan x}{x}+{\sec }^{2}x\log x+\frac{1}{2}\times \frac{2}{x^2+1}\times x \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{\tan x}{x}+{\sec }^{2}x\log x+\frac{x}{x^2+1} \\ \Rightarrow \frac{dy}{dx}=\left(x^{\tan x}+\sqrt{\frac{x^2+1}{2}}\right)\left(\frac{\tan x}{x}+{\sec }^{2}x\log x+\frac{x}{x^2+1}\right) \\ \Rightarrow \frac{dy}{dx}=x^{\tan x}\left(\frac{\tan x}{x}+{\sec }^{2}x\log x\right)+\frac{x}{\sqrt{2x^2+2}} \end{gathered}$$