Question

Find d/dx tan⁻¹ √1+x²-1/x, x ∈ R - {0}

Answer 1

Answer:

$\bf\frac{dy}{dx}=\frac{1}{2(1+x^2)}$

Step-by-step explanation:

Find d/dx tan⁻¹ √1+x²-1/x, x ∈ R - {0}

I have applied substituition method to solve this problem

$\text{Let y=}tan^{-1}(\frac{\sqrt{1+x^2}-1}{x})$

$\text{take x=}tan\theta$

$\implies\:\theta=tan^{-1}x$

$y=tan^{-1}(\frac{\sqrt{1+tan^2\theta}-1}{tan\theta})$

$y=tan^{-1}(\frac{\sqrt{sec^2\theta}-1}{tan\theta})$

$y=tan^{-1}(\frac{sec\theta-1}{tan\theta})$

$y=tan^{-1}(\frac{\frac{1}{cos\theta}-1}{\frac{sin\theta}{cos\theta}})$

$y=tan^{-1}(\frac{\frac{1-cos\theta}{cos\theta}}{\frac{sin\theta}{cos\theta}})$

$y=tan^{-1}(\frac{1-cos\theta}{sin\theta})$

using

$\boxed{1-cosA=2\:sin^2\frac{A}{2}\:\text{ and }sinA=2\:sin\frac{A}{2}\:cos\frac{A}{2}}$

$y=tan^{-1}(\frac{2\:sin^2\frac{\theta}{2}}{2\:sin\frac{\theta}{2}\:cos\frac{\theta}{2}})$

$y=tan^{-1}(\frac{sin\frac{\theta}{2}}{cos\frac{\theta}{2}})$

$y=tan^{-1}(tan\frac{\theta}{2})$

$y=\frac{\theta}{2}$

$y=\frac{1}{2}\theta$

$y=\frac{1}{2}\:tan^{-1}x$

Differentiate with respect to x

$\frac{dy}{dx}=\frac{1}{2}\:(\frac{1}{1+x^2})$

$\implies\boxed{\bf\frac{dy}{dx}=\frac{1}{2(1+x^2)}}$