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  • Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 28 sub question (i) maths

Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 28 sub question (i) maths

Answers (1)

Answer:  \frac{d y}{d x}=(\sin x)^{x}[x \cot x+\log \sin x]+\frac{1}{2 \sqrt{x-x^{2}}}

Hint: Differentiate the equation taking log on both sides

Given:

Solution:  

        \begin{aligned} &y=(\sin x)^{x}+\sin ^{-1}(\sqrt{x}) \\\\ &y=y_{1}+y_{2} \end{aligned}

Diff w.r.t x

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

                                            y_{1}=(\sin x)^{x}

Taking log on both sides

        \log y_{1}=\log (\sin x)^{x} \quad\left[\because \log m^{n}=n \log m\right]

        \log y_{1}=x \log \sin x

        \frac{1}{y_{1}} \frac{d y}{d x}=x \cdot \frac{1}{\sin x} \cdot \cos x+\log \sin x \cdot 1                        \left[\begin{array}{l} \because \frac{d}{d x} \log x=\frac{1}{x} \\\\ \frac{d}{d x} I . I I=I \frac{d}{d x} I I+I I \frac{d}{d x} I \end{array}\right]

        \begin{aligned} &\frac{d y_{1}}{d x}=y_{1}[x \cdot \cot x+\log \sin x] \\\\ &\frac{d y_{1}}{d x}=(\sin x)^{n}[x . \cot x+\log \sin x] \end{aligned}            .......(2)

        \begin{aligned} &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}) \end{aligned}

        \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}=(\sin x)^{x}[x \cot x+\log \sin x]+\frac{1}{2 \sqrt{x-x^{2}}}

 

 

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