Differentiation of $ an ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x} ight)$ w.r.t. $\cos ^{-1}\left(\sqrt{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}} ight)$, is

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Differentiation of $	an ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}ight)$ w.r.t. $\cos ^{-1}\left(\sqrt{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}}ight)$, is

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Let $\mathrm{u}=	an ^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}ight]$ and $v=\cos ^{-1}\left[\sqrt{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}}ight]$ Put $x=	an 	heta$, then $	heta=	an ^{-1} x$ $\begin{aligned} 	herefore \mathrm{u} & =	an ^{-1}\left[\frac{\sqrt{1+	an ^2 	heta}-1}{	an 	heta}ight]=	an ^{-1}\left[\frac{\sec 	heta-1}{	an 	heta}ight] \ & =	an ^{-1}\left[\frac{\frac{1}{\cos 	heta}-1}{\frac{\sin 	heta}{\cos 	heta}}ight]=	an ^{-1}\left[\frac{1-\cos 	heta}{\sin 	heta}ight] \ & =	an ^{-1}\left[\frac{2 \sin ^2 \frac{	heta}{2}}{2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}ight]=	an ^{-1}\left(	an \frac{	heta}{2}ight)=\frac{	heta}{2} \end{aligned}$ $	herefore \quad \mathrm{u}=\frac{	an ^{-1} x}{2}$ $\begin{aligned} v & =\cos ^{-1}\left[\sqrt{\frac{1+\sqrt{1+	an ^2 	heta}}{2 \sqrt{1+	an ^2 	heta}}}ight] \ & =\cos ^{-1}\left[\sqrt{\frac{1+\sec 	heta}{2 \sec 	heta}}ight] \ & =\cos ^{-1}\left[\sqrt{\frac{1+\frac{1}{\cos 	heta}}{\frac{2}{\cos 	heta}}}ight] \ & =\cos ^{-1}\left[\sqrt{\frac{1+\cos 	heta}{2}}ight] \ & =\cos ^{-1}\left(\sqrt{\frac{2 \cos ^2\left(\frac{	heta}{2}ight)}{2}}ight) \ & =\cos ^{-1}\left(\cos \frac{	heta}{2}ight)-\frac{0}{2} \end{aligned}$ $	herefore \quad v=\frac{	an ^{-1} x}{2}$ From (i) and (ii), we get $\mathrm{u}=\mathrm{v}$ $	herefore \quad \frac{d u}{d v}=1$

Differentiation of $\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ w.r.t. $\cos…

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