If the solve the problem

Basic idea:

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

The 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- $\frac{d}{d x}(u v)=u \frac{d v}{d x}+v \frac{d u}{d x}$

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

Let's solve now:

Given, $y=2 \sin x+3 \cos x \ldots . .$ equation 1

As we have to prove : $\frac{d^{2} y}{d x^{2}}+y=0$.

We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.

Let's find $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$

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.

$\therefore \frac{d y}{d x}=\frac{d}{d x}(2 \sin x+3 \cos x)=2 \frac{d}{d x}(\sin x)+3 \frac{d}{d x}(\cos x)$

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

$=2 \cos x-3 \sin x$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=2 \cos \mathrm{x}-3 \sin \mathrm{x}$

$=2 \cos x-3 \sin x$'

$\therefore \frac{d y}{d x}=2 \cos x-3 \sin x$

Differentiating again with respect to $x$ :

$\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{d}}{\mathrm{dx}}(2 \cos \mathrm{x}-3 \sin \mathrm{x})=\frac{2 \mathrm{~d}}{\mathrm{dx}} \cos \mathrm{x}-3 \frac{\mathrm{d}}{\mathrm{dx}} \sin \mathrm{x}$

$\frac{d^{2} y}{d x^{2}}=-2 \sin x-3 \cos x$

From equation 1 we have:

$y=2 \sin x+3 \cos x$

$\therefore \frac{d^{2} y}{d x^{2}}=-(2 \sin x+3 \cos x)=-y$

$\therefore \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\mathrm{y}=0 \ldots . .$ proved