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#### Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 26

Answer:  $\frac{d y}{d x}$  $=(\sin x)^{\cos x}[\cot x \cdot \cos x-\sin x \log (\sin x)]+$$(\cos x)^{\sin x}[\cos x \cdot \log (\cos x)-\sin x \tan x]$

Hint:  Differentiate the equation taking log on both sides

Given:

Solution:

\begin{aligned} &y=(\sin x)^{\cos x}+(\cos x)^{\sin x} \\\\ &y_{1}=(\sin x)^{\cos x} \text { and } \end{aligned}

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

Now   $y_{1}=(\sin x)^{\cos x}$

$\log y_{1}=\cos x \cdot \log (\sin x)$

On Diff both sides w.r.t x

$\frac{1}{y_{1}} \frac{d y_{1}}{d x}=\left[\cos x \cdot \frac{1}{\sin x} \cdot \cos x+\log (\sin x)-\sin x\right]$

$\frac{d y_{1}}{d x}=(\sin x)^{\cos x}[\cot x \cdot \cos x-\sin x \log (\sin x)]$

\begin{aligned} &y_{2}=(\cos x)^{\sin x} \\\\ &\log y_{2}=\sin \log (\cos x) \end{aligned}

On diff both sides w.r.t x

$\frac{1}{y_{2}} \frac{d y_{2}}{d x}=\left[\sin x \cdot \frac{1}{\cos x} \times-\sin x+\log (\cos x) \times \cos x\right]$

$\frac{d y_{2}}{d x}=(\cos x)^{\sin x}[-\sin x \cdot \tan x+\cos x \cdot \log (\cos x)]$

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

$=(\sin x)^{\cos x}[\cot x \cdot \cos x-\sin x \log (\sin x)]+$$(\cos x)^{\sin x}[\cos x \cdot \log (\cos x)-\sin x \tan x]$