#### Please solve RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 33 maths textbook solution

$-cot\: y.cosec^{2}y$

Hint:

You must know the derivative of cos inverse function

Given:

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

Solution:

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

\begin{aligned} &\text { Differentiate the equation w.r.t } x\\ &\frac{d y}{d x}=\frac{d\left(\cos ^{-1} x\right)}{d x}\\ &=\frac{-1}{\sqrt{1-x^{2}}}=-\left(1-x^{2}\right)^{\frac{-1}{2}}\\ &\frac{d^{2} y}{d x^{2}}=\frac{d\left[-\left(1-x^{2}\right)^{\frac{-1}{2}}\right]}{d x} \end{aligned}

\begin{aligned} &=-\left(\frac{-1}{2}\right)\left(1-x^{2}\right)^{\frac{-3}{2}} \times(-2 x) \\ &=\frac{1}{2 \sqrt{\left(1-x^{2}\right)^{3}}} \times(-2 x) \\ &\frac{d^{2} y}{d x^{2}}=\frac{-x}{\sqrt{\left(1-x^{2}\right)^{3}}} \end{aligned}

\begin{aligned} &y=\cos ^{-1} x\\ &x=\cos y\\ &\text { Put in above equation }\\ &\frac{d^{2} y}{d x^{2}}=\frac{-\cos y}{\sqrt{\left(1-\cos ^{2} y\right)^{3}}} \end{aligned}

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{-\cos y}{\sqrt{\sin ^{3} y}} \\ &=\frac{-\cos y}{\sin ^{3} y} \\ &=\frac{-\cos y}{\sin y} \times \frac{1}{\sin ^{2} y} \\ &\therefore \frac{d^{2} y}{d x^{2}}=-\cot y \cdot \operatorname{cosec}^{2} y \end{aligned}