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explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 8 maths

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Answer: \frac{2}{(2 x-3) \log 7}

Hint:  You must know the rules of solving derivative of logarithm function.

Given: \log _{7}(2 x-3)


Let   y=\log _{7}(2 x-3)

Differentiating with respect to x,

\frac{d y}{d x}=\frac{d}{d x}\left[\log _{7}(2 x-3)\right]

\frac{d y}{d x}=\frac{d}{d x}\left[\frac{\log (2 x-3)}{\log 7}\right]                        \left[\therefore \log _{a} b=\frac{\log b}{\log a}\right]

\begin{aligned} &\frac{d y}{d x}=\frac{1}{\log 7} \frac{d}{d x}[\log (2 x-3)] \\ &\frac{d y}{d x}=\frac{1}{\log 7} \times \frac{1}{(2 x-3)} \frac{d}{d x}(2 x-3) \end{aligned}          [   using chain rule ] 

\frac{d y}{d x}=\frac{2}{(2 x-3) \log 7}

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