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
$\frac{d^2y}{dx^2}=\frac{-cosy}{\sqrt{(1-cos^2y)^3}}$
$⇒\frac{d^2y}{dx^2}=\frac{-cosy}{\sqrt{(sin^2y)^3}}$
$=\frac{-cosy}{sin^3y}$
$=\frac{-cosy}{siny}\times\frac{1}{sin^2y}$
$⇒\frac{d^2y}{dx^2}=-coty.cosec^2y$