If the solve the problem

Basic idea:

$\sqrt{S e c o n d}$ order derivative is nothing but derivative of derivative i.e. $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}\right)$

$\sqrt{T h e}$ idea of chain rule of differentiation: If $\mathrm{f}$ is any real-valued function which is the composition of two functions $u$ and $v$, i.e. $f=v(u(x))$. For the sake of simplicity just assume $t=u(x)$

Then $f=v(t)$. By chain rule, we can write the derivative of $f$ w.r.t to $x$ as:

$\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dx}}$

$\sqrt{P r o d u c t}$ rule of differentiation- $\frac{d}{d x}(u v)=u \frac{d v}{d x}+v \frac{d u}{d x}$

Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let's solve now:

Given,

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

TO prove :

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

Clearly from the expression to be proved we can easily observe that we need to just find the second derivative of given function.

Given, $y=e^{x} \cos x$

We have to find $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$

$\mathrm{AS}, \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

So lets first find $d y / d x$ and differentiate it again.

$\therefore \frac{d y}{d x}=\frac{d}{d x}\left(e^{-x} \cos x\right)$

Let $u=e^{-x}$ and $v=\cos x$

As, $y=u^{*} v$

$\therefore$ using product rule of differentiation:

$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{-\mathrm{x}} \frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{x})+\cos \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}} \mathrm{e}^{-\mathrm{x}}$

$\frac{d y}{d x}=-e^{-x} \sin x-e^{-x} \cos x$

$\left[\because \frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{x})=-\sin \mathrm{x} \& \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{e}^{-\mathrm{x}}=-\mathrm{e}^{-\mathrm{x}}\right]$

Again differentiating w.r.t $\mathrm{x}$ :

$\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(-e^{-x} \sin x-e^{-x} \cos x\right)$

$=\frac{d}{d x}\left(-e^{-x} \sin x\right)-\frac{d}{d x}\left(e^{-x} \cos x\right)$

Again using the product rule :

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\mathrm{e}^{-\mathrm{x}} \frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x})-\sin \mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{e}^{-\mathrm{x}}-\mathrm{e}^{-\mathrm{x}} \frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{x})-\cos \mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{-\mathrm{x}}\right)$

$\frac{d^{2} y}{d x^{2}}=-e^{-x} \cos x+e^{-x} \sin x+e^{-x} \sin x+e^{-x} \cos x$

$\left[\because \frac{d}{d x}(\cos x)=-\sin x, \frac{d}{d x} e^{-x}=-e^{-x}\right]$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=2 \mathrm{e}^{-\mathrm{x}} \sin \mathrm{x} \ldots$ proved