Doubt from Definite integrals

Please explain why derivative of d/dx( tan inverse (1/X)) not equal to tan inverse (1/X) in interval [-1,1]

You may show this solution to Fitjee ..whether it is correct or not

\dfrac{\mathrm{d}}{\mathrm{dx}}\tan^{-1} \dfrac{1}{x} =-\dfrac{1}{1+x^2}

But antiderivative of \dfrac{1}{1+x^2} is \cot^{-1}x

So an alternative presentation of the solution would be

\tan^{-1} \dfrac{1}{x} = \cot^{-1}x ,x>0

= \cot^{-1}x - \pi, x<0

We see that \dfrac{\mathrm{d}}{\mathrm{dx}}\tan^{-1} \dfrac{1}{x}=\dfrac{\mathrm{d}}{\mathrm{dx}} \cot^{-1}x in either case

Hence \displaystyle \int_{-1}^1 \dfrac{\mathrm{d}}{\mathrm{dx}}\tan^{-1} \dfrac{1}{x} \ \mathrm{dx}= \int_{-1}^1 \dfrac{\mathrm{d}}{\mathrm{dx}} \cot^{-1}x\ \mathrm{dx}

= \displaystyle |\cot^{-1} x|_{-1}^1 = \dfrac{\pi}{4} - \dfrac{3\pi}{4}=-\dfrac{\pi}{2}

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Very nice explanation