#### Please Solve RD Sharma Class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Questions Maths Textbook Solution Question 15

(c)

Hint:

We must have known about the derivative of inverse trigonometric functions like  $\sin^{-1}x$ .

Given:

$y=\sin \left(m \sin ^{-1} x\right)$

Explanation:

$y=\sin \left(m \sin ^{-1} x\right)$

Differentiating both sides with respect $x$,

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

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

Again differentiate with respect to $x$ ,

$\frac{d^{2} y}{d x^{2}}=m\left[\left(\sqrt{1-x^{2}}\right)\left(-\sin \left(m\left(\sin ^{-1} x\right)\right) \times \frac{m}{\sqrt{1-x^{2}}}\right)-\cos \left(m\left(\sin ^{-1} x\right)\right) \times \frac{1}{2 \sqrt{1-x^{2}}} \times(0-2 x)\right] \\$

$\frac{d^{2} y}{d x^{2}}=m\left[\frac{\left(-m \sin \left(m\left(\sin ^{-1} x\right)\right)\right)+\cos \left(m\left(\sin ^{-1} x\right)\right) \times \frac{x}{\sqrt{1-x^{2}}}}{1-x^{2}}\right]$

\begin{aligned} & \\ &\frac{d^{2} y}{d x^{2}}=m\left[\frac{\left(-m \sin \left(m\left(\sin ^{-1} x\right)\right)\right)}{1-x^{2}}+\frac{x \cos \left(m\left(\sin ^{-1} x\right)\right)}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}\right] \end{aligned}

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

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

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

$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-m x \frac{d y}{d x}+m^{2} y=0$     or

\begin{aligned} & \\ &\left(1-x^{2}\right) y^{2}-m x y^{1}=-m^{2} y \end{aligned}

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