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EXERCISE 1.2 · Q58

Q.Differentiate the following w.r.t. xx: tan⁡−1(log⁡x)\tan^{-1}(\log x)

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Let u=log⁡xu=\log x, so dudx=1x\dfrac{du}{dx}=\dfrac1x. Using ddxtan⁡−1u=u′1+u2\dfrac{d}{dx}\tan^{-1}u=\dfrac{u'}{1+u^2}, we get $\dfrac{d}{dx}\tan^{-1}(\log x)=\dfrac{1/x}{1+(\log x)^2}=\dfrac{ …

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