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  • Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 28 sub question (ii) maths textbook solution

Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 28 sub question (ii) maths textbook solution

Answers (1)

Answer: \frac{d y}{d x}=x^{\sin x}\left(\frac{\sin x}{x}+\log x \cos x\right)+\frac{1}{2 \sqrt{x-x^{2}}}

Hint:  Differentiate the equation taking log on both sides

Given: y=x^{\sin x}+\sin ^{-1}(\sqrt{x})

Solution:  y=x^{\sin x}+\sin ^{-1}(\sqrt{x})

                y=y_{1}+y_{2}

Diff w.r.t x

        \frac{d y}{d x}=\frac{d y_{1}}{d x}+\frac{d y_{2}}{d x}                    .......(1)

        \begin{aligned} &\text { Let } y_{1}=x^{\sin x} \\\\ &\log y_{1}=\log x(\sin x)^{x} \end{aligned}

On diff both side with respect to x we get

\frac{1}{y} \frac{d y}{d x}=\sin x \frac{d}{d x} \log x+\log x \frac{d}{d x} \sin x

\frac{1}{y} \frac{d y}{d x}=\sin x \frac{1}{x}+\log x \cos x

\frac{d y}{d x}=y\left(\frac{\sin x}{x}+\log x \cos x\right)

\frac{d y}{d x}=x^{\sin x}\left(\frac{\sin x}{x}+\log x \cos x\right)        from 1

y_{2}=\sin ^{-1}(\sqrt{x})

\frac{d y_{2}}{d x}=\frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{d}{d x}(\sqrt{x}) \quad\left[\because \frac{d}{d x} \sin ^{-1}(\sqrt{x})=\frac{1}{\sqrt{1-x^{2}}}\right]

\begin{aligned} &\frac{d y_{2}}{d x}=\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}} \\\\ &\frac{d y_{2}}{d x}=\frac{1}{2 \sqrt{x(1-x)}} \end{aligned}

\frac{d y_{2}}{d x}=\frac{1}{2 \sqrt{x-x^{2}}}    ........(3)

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

\frac{d y}{d x}=x^{\sin x}\left(\frac{\sin x}{x}+\log x \cos x\right)+\frac{1}{2 \sqrt{x-x^{2}}}

 

 

 

Posted by

infoexpert26

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