If the solve the problem


If $y=\frac{\log x}{x}$, show that $\frac{d^{2} y}{d x^{2}}=\frac{2 \log x-3}{x^{3}}$


Basic 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:

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

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, $y=\frac{\log x}{x} \ldots . .$ equation 1

As we have to prove : $\frac{d^{2} y}{d x^{2}}=\frac{2 \log x-3}{x^{3}} . .$

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 $\frac{d^{2} y}{d x^{2}}$

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.

As $y$ is the product of two functions $u$ and $v$

Let $u=\log x$ and $v=1 / x$

Using product rule of differentiation:

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

$\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\log \mathrm{x}}{\mathrm{x}}\right)=\log \mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}} \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}} \frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{x}$

$\frac{d y}{d x}=-\frac{1}{x^{2}} \log x+\frac{1}{x^{2}}$

$\frac{d y}{d x}=\frac{1}{x^{2}}(1-\log x)$

Again using the product rule to find $\frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}$ :

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=(1-\log \mathrm{x}) \frac{\mathrm{d}}{\mathrm{dx}} \frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{x}^{2}} \frac{\mathrm{d}}{\mathrm{dx}}(1-\log \mathrm{x})$

$\left[\because \frac{\mathrm{d}}{\mathrm{dx}}(\log x)=\frac{1}{x} \& \frac{\mathrm{d}}{\mathrm{dx}}\left(x^{\mathrm{n}}\right)=n x^{\mathrm{n}-1}\right]$

$=-2\left(\frac{1-\log x}{x^{3}}\right)-\frac{1}{x^{3}}$

$\therefore \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{2 \log \mathrm{x}-3}{\mathrm{x}^{2}} \ldots$ proved

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