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Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 22

Answers (1)

Answer:  \frac{d y}{d x}=\frac{e^{a x} \sec x \log x}{\sqrt{1-2 x}}\left[a+\tan x+\frac{1}{x \log x}+\frac{1}{1-2 x}\right]

Hint:  Differentiate the equation taking log on both sides

Given:  y=\frac{e^{a x} \sec x \log x}{\sqrt{1-2 x}}

Solution:  

        y=\frac{e^{a x} \sec x \log x}{\sqrt{1-2 x}}

Taking log on both sides,

        \log y=\log \left[\frac{e^{a x} \sec x \log x}{\sqrt{1-2 x}}\right]

                    =\log \left(e^{a x}\right)+\log (\sec x)+\log (\log x)-\log (1-2 x)^{\frac{1}{2}}

        \log y=a x+\log (\sec x)+\log (\log x)-\frac{1}{2} \log (1-2 x)

        \frac{1}{y} \frac{d y}{d x}=a+\frac{\sec x \cdot \tan x}{\sec x}+\frac{1}{\log x} \cdot \frac{1}{x}-\frac{1}{2} \cdot \frac{1(-2)}{(1-2 x)}

        \frac{d y}{d x}=y\left[a+\tan x+\frac{1}{x \log x}+\frac{1}{1-2 x}\right]

        \frac{d y}{d x}=\frac{e^{a x} \sec x \log x}{\sqrt{1-2 x}}\left[a+\tan x+\frac{1}{x \log x}+\frac{1}{1-2 x}\right]

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