Question

Find $\frac{dy}{dx}$

$y={\left(\tan x\right)}^{\cot x}+{\left(\cot x\right)}^{\tan x}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y={\left(\tan x\right)}^{\cot x}+{\left(\cot x\right)}^{\tan x} \\ \Rightarrow y=e^{\log {\left(\tan x\right)}^{\cot x}}+e^{\log {\left(\cot x\right)}^{\tan x}} \\ \Rightarrow y=e^{\cot x\log \tan x}+e^{\tan x\log \left(\cot x\right)} \end{gathered}$$
Differentiating with respect to x using chain rule and product rule,
$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(e^{\cot x\log \tan x}\right)+\frac{d}{dx}\left(e^{\tan x\mathrm{logcot}x}\right) \\ =e^{\cot x\log \tan x}\frac{d}{dx}\left({}^{\cot x\log \tan x}\right)+e^{\tan x\mathrm{logcot}x}\frac{d}{dx}\left({}^{\tan x\mathrm{logcot}x}\right) \\ =e^{\log {\left(\tan x\right)}^{\cot x}}\left[\cot x\frac{d}{dx}\left(\log \tan x\right)+\log \tan x\frac{d}{dx}\left(\cot x\right)\right]+e^{\log \left(\cot x\right)\tan x}\left[\tan x\frac{d}{dx}\left(\log \cot x\right)+\mathrm{logcot}x\frac{d}{dx}\left(\tan x\right)\right] \\ ={\left(\tan x\right)}^{\cot x}\left[\cot x\times \left(\frac{1}{\tan x}\right)\frac{d}{dx}\left(\tan x\right)+\log \tan x\left(-{\mathrm{cosec}}^{2}x\right)\right]+{\left(\cot x\right)}^{\tan x}\left[\tan x\times \left(\frac{1}{\cot x}\right)\frac{d}{dx}\left(\cot x\right)+\log \cot x\left({\sec }^2\mathrm{x}\right)\right] \\ ={\left(\tan x\right)}^{\cot x}\left[\left(\frac{{\mathrm{cosec}}^{2}x}{{\sec }^{2}x}\right)\left({\sec }^{2}x\right)-{\mathrm{cosec}}^{2}x\log \tan x\right]+{\left(\cot x\right)}^{\tan x}\left[\left(\frac{{\sec }^{2}x}{{\mathrm{cosec}}^{2}x}\right)\left(-{\mathrm{cosec}}^{2}x\right)+{\sec }^{2}x\log \cot x\right] \\ ={\left(\tan x\right)}^{\cot x}\left[{\mathrm{cosec}}^{2}x-{\mathrm{cosec}}^{2}x\log \tan x\right]+{\left(\mathrm{cotx}\right)}^{\tan x}\left[{\sec }^{2}x\log \cot x-{\sec }^{2}x\right] \\ ={\left(\tan x\right)}^{\cot x}{\mathrm{cosec}}^{2}x\left[1-\log \tan x\right]+{\left(\mathrm{cotx}\right)}^{\tan x}{\sec }^{2}x\left[\log \cot x-1\right] \end{gathered}$$