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#### Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 29 sub question (ii)

Answer:  $\frac{d y}{d x}=x^{x}(1+\log x)+(\sin x)^{x}[x \cot x+\log (\sin x)]$

Hint: Differentiate the equation taking log on both sides

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

Solution:  $y=x^{x}+(\sin x)^{x}$

$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$    .............(1)

\begin{aligned} &\text { Let } u=x^{x} \\ &\log u=\log x^{x} \\ &\log u=x \log x \end{aligned}

Diff w.r.t x

\begin{aligned} &\frac{1}{u} \frac{d u}{d x}=x \cdot \frac{1}{x}+\log x \\\\ &\frac{d u}{d x}=u(1+\log x) \\\\ &\frac{d u}{d x}=x^{x}(1+\log x) \end{aligned}   ..............(2)

\begin{aligned} &\text { Again let } v=(\sin x)^{x} \\\\ &\log v=\log (\sin x)^{x} \\\\ &\log v=x \log (\sin x) \end{aligned}

Diff w.r.t x

\begin{aligned} &\frac{1}{v} \frac{d v}{d x}=x \cdot \frac{\cos x}{\sin x}+\log (\sin x) \\\\ &\frac{d v}{d x}=v[x \cdot \cot x+\log (\sin x)] \\\\ &\frac{d v}{d x}=(\sin x)^{x}[x \cdot \cot x+\log (\sin x)] \end{aligned}    ...........(3)

Put (2) and (3) in eq (1)

$\frac{d y}{d x}=x^{x}(1+\log x)+(\sin x)^{x}[x \cot x+\log (\sin x)]$