#### Provide solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 2

$2e^{-x}sin\: x$

Hint:

You have to show

$\frac{d^{2}y}{dx^{2}}=2e^{-x}sin\: x$

Given:

If y=e-xcos x, show that

$\frac{d^{2}y}{dx^{2}}=2e^{-x}sin\: x$

Solution:

Let y=e-xcos x

Use multiplicative rule

\begin{aligned}\text { As } U V=U V^{1}+U^{1} V \\ \text { Where } U=e^{-x} \& V=\cos x \\\frac{d y}{d x}=e^{-x} \frac{d}{d x} \cos x+\cos x \frac{d}{d x} e^{-x}\\ \frac{d y}{d x}=-e^{-x} \sin x-\cos x e^{-x} & \left(\frac{d \cos x}{d x}=-\sin x, \frac{d e^{-x}}{d x}=-1 e^{-x}\right) \end{aligned}

Again differentiating w.r.t.x we get

Use multiplicative rule

\begin{aligned} &\text { As } U V=U V^{1}+U^{1} V \\ &\text { Where } U=\cos x\: \&\: V=e^{-x} \\ &\frac{d^{2} y}{d x^{2}}=-\left[\sin x \cdot \frac{d}{d x} e^{-x}+e^{-x} \cdot \frac{d}{d x} \sin x+\cos x \frac{d}{d x} e^{-x}+e^{-x} \frac{d}{d x} \cos x\right] \\ &\frac{d^{2} y}{d x^{2}}=-\left[\sin x \cdot\left(-e^{-x}\right)+e^{-x} \cos x+\cos x\left(-e^{-x}\right)+e^{-x}(-\sin x)\right] \end{aligned}

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

Hence proved