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

Question:

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

$x \cos x$

Solution:

$\sqrt{B a s i c}$ 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 $f$ 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, $y=x \cos x$

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

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.

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x} \cos \mathrm{x})$

Let $u=x$ and $v=\cos x$

As, $y=U V$

$\therefore$ Using product rule of differentiation:

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{X} \frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{X})+\cos \mathrm{X} \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{X}$

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

$\left[\because \frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{x})=-\sin \mathrm{x}\right.$ and $\left.\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1}\right]$

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

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

$=\frac{\mathrm{d}}{\mathrm{dx}}(-\mathrm{x} \sin \mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}$

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

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

$\left[\because \frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x})=\cos \mathrm{x}\right.$ and $\left.\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1}\right]$

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

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