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

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

Answers (1)

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)]

 

 

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