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explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 66 maths

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Answer: Proved

Hint: you must know the rules of derivative of logarithm functions.

Given:  \mathrm{y}=(\mathrm{x}-1) \log (x-1)-(x+1) \log (x+1)

Prove: \frac{d y}{d x}=\log \left(\frac{x-1}{1+x}\right)

Solution:

Let  \mathrm{y}=(\mathrm{x}-1) \log (x-1)-(x+1) \log (x+1)

Differentiate with respect to x, use product rule

\frac{d y}{d x}=\frac{d}{d x}[(\mathrm{x}-1) \log (x-1)-(x+1) \log (x+1)]

\frac{d y}{d x}=\left[(x-1) \times \frac{1}{(x-1)}+\log (x-1)\right]-\left[(x+1) \times \frac{1}{(x+1)}+\log (x+1)\right]   [ use product rule]         

\frac{d y}{d x}=[1+\log (x-1)]-[1+\log (x+1)]

\frac{d y}{d x}=\log \left(\frac{x-1}{1+x}\right)

∴ Proved

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