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

Question:

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

$e^{6 x} \cos 3 x$

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:

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

$\sqrt{P r o d u c t}$ rule of differentiation- $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{uv})=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}$

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

Let's solve now:

Given, $y=e^{6 x} \cos 3 x$

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

$A S, \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)$

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

$\therefore \frac{d y}{d x}=\frac{d}{d x}\left(e^{6 x} \cos 3 x\right)$

Let $u=e^{6 x}$ and $v=\cos 3 x$

As, $y=U V$

$\therefore$ Using product rule of differentiation:

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

$\therefore \frac{d y}{d x}=e^{6 x} \frac{d}{d x}(\cos 3 x)+\cos 3 x \frac{d}{d x} e^{6 x}$

$\frac{d y}{d x}=-3 e^{6 x} \sin 3 x+6 e^{6 x} \cos 3 x\left[\because \frac{d}{d x}(\cos a x)=-a \sin a x, a\right.$ is any constant $\left.\& \frac{d}{d x} e^{a x}=a e^{x}\right]$

Again differentiating w.r.t x:

$\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(-3 e^{6 x} \sin 3 x+6 e^{6 x} \cos 3 x\right)$

$=\frac{d}{d x}\left(-3 e^{6 x} \sin 3 x\right)+\frac{d}{d x}\left(6 e^{6 x} \cos 3 x\right)$

Again using the product rule :

$\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(-3 e^{6 x} \sin 3 x+6 e^{6 x} \cos 3 x\right)$

$=\frac{d}{d x}\left(-3 e^{6 x} \sin 3 x\right)+\frac{d}{d x}\left(6 e^{6 x} \cos 3 x\right)$

Again using the product rule :

$\frac{d^{2} y}{d x^{2}}=-3 e^{6 x} \frac{d}{d x}(\sin 3 x)-3 \sin 3 x \frac{d}{d x} e^{6 x}+6 e^{6 x} \frac{d}{d x}(\cos 3 x)+\cos 3 x \frac{d}{d x}\left(6 e^{6 x}\right)$

$\frac{d^{2} y}{d x^{2}}=-9 e^{6 x} \cos 3 x-18 e^{6 x} \sin 3 x-18 e^{6 x} \sin 3 x+36 e^{6 x} \cos 3 x$

$\frac{d^{2} y}{d x^{2}}=27 e^{6 x} \cos 3 x-36 e^{6 x} \sin 3 x$