If the solve the problem

Basic idea:

Second order derivative is nothing but derivative of derivative i.e. $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)$

$\sqrt{T h e}$ 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:

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

Product rule of differentiation- $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{uv})=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}$

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 :

$\frac{d^{2} y}{d x^{2}}=2 e^{x} \cos \left(x+\frac{\pi}{2}\right)$

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 $\frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}$

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

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

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

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

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

$\therefore$ Using product rule of differentiation:

$\therefore \frac{d y}{d x}=e^{x} \frac{d}{d x}(\cos x)+\cos x \frac{d y}{d x} e^{x}$

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

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

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

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

Again using the product rule :

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

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

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

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-2 \mathrm{e}^{\mathrm{x}} \sin \mathrm{x}[\because-\sin x=\cos (x+\pi / 2)]$

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