# If findin terms of y alone.

Question:

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

Solution:

It is given that, $y=\cos ^{-1} x$

Then,

$\frac{d y}{d x}=\frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^{2}}}=-\left(1-x^{2}\right)^{\frac{-1}{2}}$

$\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left[-\left(1-x^{2}\right)^{\frac{-1}{2}}\right]$

$=-\left(-\frac{1}{2}\right) \cdot\left(1-x^{2}\right)^{\frac{-3}{2}} \cdot \frac{d}{d x}\left(1-x^{2}\right)$

$=\frac{1}{2 \sqrt{\left(1-x^{2}\right)^{3}}} \times(-2 x)$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{-x}{\sqrt{\left(1-x^{2}\right)^{3}}}$   ...(1)

$y=\cos ^{-1} x \Rightarrow x=\cos y$

Putting $x=\cos y$ in equation (i), we obtain

$\frac{d^{2} y}{d x^{2}}=\frac{-\cos y}{\sqrt{\left(1-\cos ^{2} y\right)^{3}}}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{-\cos y}{\sqrt{\left(\sin ^{2} y\right)^{3}}}$

$=\frac{-\cos y}{\sin ^{3} y}$

$=\frac{-\cos y}{\sin y} \times \frac{1}{\sin ^{2} y}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=-\cot y \cdot \operatorname{cosec}^{2} y$