Question

Differentiate

${\cos }^{-1}\left\{\frac{x}{\sqrt{x^2+a^2}}\right\}$

Answer 1

$$\begin{gathered} \mathrm{Let},y={\cos }^{-1}\left\{\frac{x}{\sqrt{x^2+a^2}}\right\} \\ \mathrm{Put}x=acot\theta \\ \Rightarrow y={\cos }^{-1}\left\{\frac{acot\theta }{\sqrt{a^{2}co{t}^2\theta +a^2}}\right\} \\ \Rightarrow y={\cos }^{-1}\left\{\frac{acot\theta }{\sqrt{a^2\left(co{t}^2\theta +1\right)}}\right\} \\ \Rightarrow y={\sin }^{-1}\left(\frac{acot\theta }{a\cos ec\theta }\right) \\ \Rightarrow y={\cos }^{-1}\left(\frac{{\displaystyle \frac{\cos \theta }{\sin \theta }}}{{\displaystyle \frac{1}{\sin \theta }}}\right) \\ \Rightarrow y={\cos }^{-1}\left(\cos \theta \right) \\ \Rightarrow y=\theta \\ \Rightarrow y=co{t}^{-1}\left(\frac{x}{a}\right)\left[\sin ce,x=acot\theta \right] \\ \mathrm{Differentiating}\mathrm{it}\mathrm{with}\mathrm{respect}\mathrm{to}x\mathrm{using}\mathrm{chain}\mathrm{rule}, \\ \frac{dy}{dx}=\frac{-1}{1+{\left({\displaystyle \frac{x}{a}}\right)}^2}\frac{d}{dx}\left(\frac{x}{a}\right) \\ \Rightarrow \frac{dy}{dx}=\frac{-a^2}{a^2+x^2}\times \left(\frac{1}{a}\right) \\ \therefore \frac{dy}{dx}=\frac{-a}{a^2+x^2} \end{gathered}$$