# Solve this

Question:

Find $\frac{d y}{d x}$, when

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

Solution:

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

$\Rightarrow y=a+b$

where $a=(\tan x)^{\cot x} ; b=(\cot x)^{\tan x}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{da}}{\mathrm{dx}}+\frac{\mathrm{db}}{\mathrm{dx}}$

$\left\{\right.$ Using chain rule, $\frac{\mathrm{d}(\mathrm{u}+\mathrm{a})}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dx}}+\frac{\mathrm{da}}{\mathrm{dx}}$ where a and $\mathrm{u}$ are any variables $\}$

$a=(\tan x)^{\cot x}$

Taking log both the sides:

$\Rightarrow \log a=\log (\tan x)^{\cot x}$

$\Rightarrow \log a=\cot x \log (\tan x)$

$\left\{\log x^{a}=\operatorname{alog} x\right\}$

Differentiating with respect to $\mathrm{x}$ :

$\Rightarrow \frac{d(\log a)}{d x}=\frac{d(\cot x \log (\tan x))}{d x}$

$\Rightarrow \frac{d(\log a)}{d x}=\cot x \times \frac{d(\log (\tan x))}{d x}+\log (\tan x) \times \frac{d(\cot x)}{d x}$

$\left\{\right.$ Using product rule, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{a} \frac{d a}{d x}=\cot x \times \frac{1}{\tan x} \frac{d(\tan x)}{d x}+\log (\tan x)\left(-\operatorname{cosec}^{2} x\right)$

$\left\{\frac{d(\log u)}{d x}=\frac{1}{u} \frac{d u}{d x} ; \frac{d(\tan x)}{d x}=\sec ^{2} x ; \frac{d(\cot x)}{d x}=-\operatorname{cosec}^{2} x\right\}$

$\Rightarrow \frac{1}{a} \frac{d a}{d x}=\cot ^{2} x\left(\sec ^{2} x\right)-\operatorname{cosec}^{2} x \log (\tan x)$

$\left\{\tan x=\frac{1}{\cot x}\right\}$

$\Rightarrow \frac{d a}{d x}=a\left\{\cot ^{2} x \sec ^{2} x-\operatorname{cosec}^{2} x \log (\tan x)\right\}$

Put the value of $a=(\tan x)^{\cot x}$;

$\Rightarrow \frac{d a}{d x}=(\tan x)^{\cot x}\left\{\cot ^{2} x \sec ^{2} x-\operatorname{cosec}^{2} x \log (\tan x)\right\}$

$b=(\cot x)^{\tan x}$

Taking log both the sides:

$\Rightarrow \log b=\log (\cot x)^{\tan x}$

$\Rightarrow \log b=\tan x \log (\cot x)$

$\left\{\log x^{a}=\operatorname{alog} x\right\}$

Differentiating with respect to $x$ :

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{b})}{\mathrm{dx}}=\frac{\mathrm{d}(\tan \mathrm{x} \log (\cot \mathrm{x}))}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{b})}{\mathrm{dx}}=\tan \mathrm{x} \times \frac{\mathrm{d}(\log (\cot \mathrm{x}))}{\mathrm{dx}}+\log (\cot \mathrm{x}) \times \frac{\mathrm{d}(\tan \mathrm{x})}{\mathrm{dx}}$

$\left\{\right.$ Using product rule, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{\mathrm{~b}} \frac{\mathrm{db}}{\mathrm{dx}}=\tan \mathrm{x} \times \frac{1}{\cot \mathrm{x}} \frac{\mathrm{d}(\cot \mathrm{x})}{\mathrm{dx}}+\log (\cot \mathrm{x})\left\{\sec ^{2} \mathrm{x}\right\}$

$\left\{\frac{d(\log u)}{d x}=\frac{1}{u} \frac{d u}{d x} ; \frac{d(\tan x)}{d x}=\sec ^{2} x ; \frac{d(\cot x)}{d x}=-\operatorname{cosec}^{2} x\right\}$

$\Rightarrow \frac{1}{\mathrm{~b}} \frac{\mathrm{db}}{\mathrm{dx}}=\tan ^{2} \mathrm{x}\left(-\operatorname{cosec}^{2} \mathrm{x}\right)+\sec ^{2} \mathrm{x} \log (\cot \mathrm{x})$

$\left\{\cot x=\frac{1}{\tan x}\right\}$

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=\mathrm{b}\left\{-\tan ^{2} \mathrm{x} \operatorname{cosec}^{2} \mathrm{x}+\sec ^{2} \mathrm{x} \log (\cot \mathrm{x})\right\}$

Put the value of $b=(\cot x)^{\tan x}$ :

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=(\cot \mathrm{x})^{\tan \mathrm{x}}\left\{\sec ^{2} \mathrm{x} \log (\cot \mathrm{x})-\tan ^{2} \mathrm{x} \operatorname{cosec}^{2} \mathrm{x}\right\}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{da}}{\mathrm{dx}}+\frac{\mathrm{db}}{\mathrm{dx}}$

$\Rightarrow \frac{d y}{d x}=(\tan x)^{\cot x}\left\{\cot ^{2} x \sec ^{2} x-\operatorname{cosec}^{2} x \log (\tan x)\right\}$

$+(\cot x)^{\tan x}\left\{\sec ^{2} x \log (\cot x)-\tan ^{2} x \operatorname{cosec}^{2} x\right\}$