Question

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

Answer 1

$y={\cos }^{-1}x$
$\Rightarrow \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}=-(1-x^2{)}^{\frac{-1}{2}}$
$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}[-(1-x^2{)}^{\frac{-1}{2}}]$
$=-(-\frac{1}{2})(1-x^2{)}^{\frac{-3}{2}}.\frac{d}{dx}(1-x^2)$
= $\frac{1}{2\sqrt{1-x^2{)}^3}}\times \left(2x\right)$
$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=\frac{-x}{2\sqrt{(1-x^2{)}^3}}$ .....(i)
Now $y={\cos }^{-1}x\Rightarrow x=\cos y$
Putting $x=\cos y$ in equation (i), we obtain
$\frac{d^{2}y}{d{x}^2}=\frac{-\cos y}{\sqrt{(1-{\cos }^{2}y{)}^3}}$
$=\frac{-\cos y}{{\sin }^{3}y}$
$=\frac{-\cos y}{\sin y}\times \frac{1}{{\sin }^{2}y}$
$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=-\cot y.co{\sec }^{2}y$