Differentiate $\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x}$ w.r.t. $\tan^{-1}\dfrac{2x\sqrt{1-x^2}}{1-2x^2}$.

Differentiate tan⁻¹dfrac√(1+x²)-1x w.r.t. tan⁻¹dfrac2x√(1-x²)1-2x².

EXERCISE 1.4 · Q179

\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x}

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Let $u=\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x}$ , $v=\tan^{-1}\dfrac{2x\sqrt{1-x^2}}{1-2x^2}$ .

Step 1 (for u). Put $x=\tan\theta$ , $\theta\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ . Then $\sqrt{1+x^2}=\sec\theta$ , so

\frac{\sqrt{1+x^2}-1}{x}=\frac{\sec\theta-1}{\tan\theta}=\frac{\dfrac{1-\cos\theta}{\cos\theta}}{\dfrac{\sin\theta}{\cos\theta}}=\frac{1-\cos\theta}{\sin\theta}=\tan\frac{\theta}{2}

(using the identity $\dfrac{1-\cos\theta}{\sin\theta}=\tan\dfrac{\theta}{2}$ ). So

$u=\tan^{-1}\left(\tan\frac{\theta}{2}\right)=\frac{\theta}{2}=\frac12\tan^{-1}x \quad\Longrightarrow\quad \frac{du}{dx}=\frac{1}{2(1+x^2)}$ …