#### Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 27

Answer:  $\frac{d y}{d x}=(\tan x)^{\cot x}\left[\operatorname{cosec}^{2} x(1-\log \tan x)\right]+(\cot x)^{\tan x}\left[\sec ^{2} x \cdot \log (\cot x)-1\right]$

Hint:  Differentiate the equation taking log on both sides

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

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

Let's assume  $y_{1}=(\tan x)^{\cot x} \text { and }$

\begin{aligned} &y_{2}=(\cot x)^{\tan x} \\\\ &\frac{d y}{d x}=\frac{d y_{1}}{d x}+\frac{d y_{2}}{d x} \end{aligned}        .......(1)

Now     $y_{1}=(\tan x)^{\cot x}$

$\log y_{1}=\cot x \cdot \log (\tan x)$

On diff both sides w.r.t. x

$\frac{1}{y_{1}} \frac{d y_{1}}{d x}=\left[\cot x \cdot \frac{1}{\tan x} \cdot \sec ^{2} x+\log (\tan x) x-\cos e c^{2} x\right]$

$\frac{d y_{1}}{d x}=(\tan x)^{\cot x}\left[\operatorname{cosec}^{2} x(1-\log (\tan x)]\right.$        .......(2)

\begin{aligned} &y_{2}=(\cot x)^{\tan x} \\\\ &\log y_{2}=\tan x \log (\cot x) \end{aligned}

On diff both sides w.r.t. x

$\frac{1}{y_{2}} \frac{d y_{2}}{d x}=\left[\tan x \cdot \frac{1}{\cot x} \cdot-\left(\cos e c^{2 x}\right)+\log \cot x \cdot \sec ^{2} x\right]$

$\frac{d y_{2}}{d x}=y_{2}\left[-\sec ^{2} x+\log (\cot x) \cdot \sec ^{2} x\right]$

$\frac{d y_{2}}{d x}=(\cot x)^{\tan x}\left[\sec ^{2} x \log (\cot x-1)\right]$            ........(3)

$\frac{d y}{d x}=\frac{d y_{1}}{d x}+\frac{d y_{2}}{d x}$

$\frac{d y}{d x}=(\tan x)^{\cot x}\left[\operatorname{cosec}^{2} x(1-\log (\tan x)]+(\cot x)^{\tan x}\left[\sec ^{2} x \log (\cot x-1)\right]\right.$