Question

Find $\frac{dy}{dx}$

$y={\left(\tan x\right)}^{\log x}+{\cos }^2\left(\frac{\pi }{4}\right)$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y={\left(\tan x\right)}^{\log x}+{\cos }^2\left(\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow y=e^{\log {\left(\tan x\right)}^{\log x}}+{\cos }^2\left(\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow y=e^{\log x\log \tan x}+{\cos }^2\left(\frac{\mathrm{\pi }}{4}\right) \end{gathered}$$
Differentiating with respect to x using chain rule,
$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(e^{\log x\log \tan x}\right)+\frac{d}{dx}{\cos }^2\left(\frac{\mathrm{\pi }}{4}\right) \\ =e^{\log x\log \tan x}\frac{d}{dx}\left(\log x\log \tan x\right)+0 \\ =e^{\log {\left(\tan x\right)}^{\log x}}\left[\log x\frac{d}{dx}\left(\log \tan x\right)+\log \tan x\frac{d}{dx}\left(\log x\right)\right] \\ ={\left(\tan x\right)}^{\log x}\left[\log x\left(\frac{1}{\tan x}\right)\frac{d}{dx}\left(\tan x\right)+\log \tan x\left(\frac{1}{x}\right)\right] \\ ={\left(\tan x\right)}^{\log x}\left[\log x\left(\frac{1}{\tan x}\right)\left({\sec }^{2}x\right)+\frac{\log \tan x}{x}\right] \\ ={\left(\tan x\right)}^{\log x}\left[\log x\left(\frac{{\sec }^{2}x}{\tan x}\right)+\frac{\log \tan x}{x}\right] \end{gathered}$$