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Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 9 maths textbook solution

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Answer:  (\sin x)^{\log x} \frac{\log (\sin x)}{x}+\log x+\sin x

Hint: Diff by \sin x^{x}

Given: (\sin x)^{\log x}


         Let y=(\sin x)^{\log x}            ........(i)

taking log on both sides,

\begin{aligned} &\log y=\log (\sin x)^{\log x} \\\\ &\log y=\log x \log (\sin x) \end{aligned}            \left[\text { Using } \log a^{b}=b \log a\right]

Differentiate w.r.t x, using product rule and chain rule,

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

                =\log _{x}\left(\frac{1}{\sin x}\right) \frac{d}{d x}(\sin x)+\log \sin x\left(\frac{1}{x}\right)

                =\frac{\log x}{\sin x} \times \cos x+\frac{\log \sin x}{x}

        \begin{aligned} \frac{1}{y} \frac{d y}{d x} &=\log x \cot x+\frac{\log \sin x}{x} \\\\ \frac{d y}{d x} &=y\left[\log x \cot x+\frac{\log \sin x}{x}\right] \\\\ \frac{d y}{d x} &=(\sin x)^{\log x}\left[\log x \cot x+\frac{\log \sin x}{x}\right] \end{aligned}                [Using equaton (i)]

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