Find the second order derivatives of each of the following functions:

Solution:

$\sqrt{B a s i c}$ Idea: Second order derivative is nothing but derivative of derivative i.e. $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

$\sqrt{T h e}$ 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:

$\frac{d f}{d x}=\frac{d v}{d t} \times \frac{d t}{d x}$

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

Let's solve now:

Given, $y=e^{x} \sin 5 x$

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.

$\therefore \frac{d y}{d x}=\frac{d}{d x}\left(e^{x} \sin 5 x\right)$

Let $u=e^{x}$ and $v=\sin 5 x$

As, $y=u v$

$\therefore$ Using product rule of differentiation:

$\frac{d y}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}$

$\therefore \frac{d y}{d x}=e^{x} \frac{d}{d x}(\sin 5 x)+\sin 5 x \frac{d}{d x} e^{x}$

$\frac{d y}{d x}=5 e^{x} \cos 5 x+e^{x} \sin 5 x$

$\left[\because \frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{ax})=\mathrm{a} \cos \mathrm{ax}\right.$, where a is any constant $\left.\& \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{e}^{\mathrm{x}}=\mathrm{e}^{\mathrm{x}}\right]$

Again differentiating w.r.t $x$ :

$\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(5 e^{x} \cos 5 x+e^{x} \sin 5 x\right)$

$=\frac{d}{d x}\left(5 e^{x} \cos 5 x\right)+\frac{d}{d x}\left(e^{x} \sin 5 x\right)$

Again using the product rule :

$\frac{d^{2} y}{d x^{2}}=e^{x} \frac{d}{d x}(\sin 5 x)+\sin 5 x \frac{d}{d x} e^{x}+5 e^{x} \frac{d}{d x}(\cos 5 x)+\cos 5 x \frac{d}{d x}\left(5 e^{x}\right)$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=5 \mathrm{e}^{\mathrm{x}} \cos 5 \mathrm{x}-25 \mathrm{e}^{\mathrm{x}} \sin 5 \mathrm{x}+\mathrm{e}^{\mathrm{x}} \sin 5 \mathrm{x}+5 \mathrm{e}^{\mathrm{x}} \cos 5 \mathrm{x}\left[\because \frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{ax})=-\mathrm{a} \sin \mathrm{ax}, \mathrm{a}\right.$ is any constant $]$

$\frac{d^{2} y}{d x^{2}}=10 e^{x} \cos 5 x-24 e^{x} \sin 5 x$