Question

If \(y=cos^{-1}x\),find \(\frac{d^2y}{dx^2}\) in terms of \(y\) alone.

Answer 1

The correct answer is \(⇒\frac{d^2y}{dx^2}=-coty.cosec^2y\)
It is given that,\(y=cos^{-1}x\)
Then,
\(\frac{dy}{dx}=\frac{d}{dx}(cos^{-1}x)=\frac{-1}{\sqrt{1-x^2}}=-(1-x^2)^{\frac{-1}{2}}\)
\(\frac{d^2y}{dx^2}=\frac{d}{dx}[-(1-x^2)^{\frac{-1}{2}}]\)
\(=-(\frac{-1}{2}).(1-x^2)^{\frac{-3}{2}}.\frac{d}{dx}(1-x^2)\)
\(=\frac{1}{2\sqrt{(1-x^2)^3}}\times (-2x)\)
\(⇒\frac{d^2y}{dx^2}=\frac{-x}{\sqrt{(1-x^2)^3}}   ...(i)\)
\(y=cos^{-1}x\)
\(⇒x=cosy\)
Putting \(x=cosy\) in equation \((i)\),we obtain
\(\frac{d^2y}{dx^2}=\frac{-cosy}{\sqrt{(1-cos^2y)^3}}\)
\(⇒\frac{d^2y}{dx^2}=\frac{-cosy}{\sqrt{(sin^2y)^3}}\)
\(=\frac{-cosy}{sin^3y}\)
\(=\frac{-cosy}{siny}\times\frac{1}{sin^2y}\)
\(⇒\frac{d^2y}{dx^2}=-coty.cosec^2y\)