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  • Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 6

Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 6

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Answer: (\log x)^{\cos x}(1+\log x \log \log x)

Hint: Diff. by applying x \log x

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

Solution:  Let y=(\log x)^{\cos x}

Taking log both sides

        \log y=\cos x \cdot \log (\log x)            \left[a s \log a^{b}=b \log a\right]

Differentiate w.r.t x

        \frac{1}{y} \frac{d y}{d x}=d \frac{(\cos x \cdot \log (\log x)}{d x}

Using product of rule \cos x \cdot \log (\log x)

        \begin{aligned} &(u v)^{\prime}=u^{\prime} v+v^{\prime} u \\\\ &\frac{1}{y} \frac{d y}{d x}=\frac{d(\cos x)}{d x} \log (\log x)+d\left(\frac{\log (\log x)}{d x}\right) \cos x \end{aligned}

        \begin{aligned} &\frac{1}{y} \frac{d y}{d x}=-\sin x \cdot \log (\log x) \cos x \\\\ &\frac{1}{y} \frac{d y}{d x}=\sin x \cdot \log (\log x)+\frac{1}{\log x} \cdot \frac{1}{x} \cos x \end{aligned}

        \begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\sin x \cdot \log (\log x)+\frac{\cos x}{x \log x} \\\\ &\frac{d y}{d x}=y\left(-\sin x \cdot \log (\log x)+\frac{\cos x}{x \log x}\right) \end{aligned}

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

        \frac{d y}{d x}=(\log x)^{\cos x}\left ( \frac{\cos x}{x\log\: x} -sin \: x.\log\left ( \log x \right ) \right )

 

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