# Find yʹ and yʹʹ. y=xln(x)

In this question, we have to find the first and second derivatives of the given function y=x ln(x)

The basic concept behind this question is the knowledge of derivatives and the rules such as the product rule of derivatives and the quotient rule of derivatives.

Given function:

$y=x \ln{\ (x)}$

For first derivative, take derivative with respect to x on both sides. We get:

$\frac{dy}{dx}=\frac{d}{dx}\ \left[x\ \ln{\ (x)}\right]$

$\frac{dy}{dx}=\frac{d}{dx}[ x\ ] ln(x)+ x\frac{d}{dx} [ln(x)]$

$\frac{dy}{dx}=\ 1 \ln{(x)}+ x\ \frac{1}{x}\$

$\frac{dy}{dx}= \ln{(x)}+ 1$

So the first derivative is:

$\frac{dy}{dx}= \ln{(x)}+ 1$

To find the second derivative, we will take the derivative of the first derivative with respect to $x$ on both sides again.

$\frac{d}{ dx}\left(\frac{dy}{dx}\right)\ =\frac{d}{dx}\ \left(\ln{(x)\ +\ 1} \right)$

$\frac{{d\ }^2y}{{dx}^2}\ =\frac{d}{dx}\ \left(\ln(x)\right) +\frac{d}{dx}\ \left(1 \right)$

$\frac{{d\ }^2y}{{dx}^2}\ =\frac{1}{x}\ + 0$

$\frac{{d\ }^2y}{{dx}^2}\ =\frac{1}{x}$

The second derivative of the function is:

$\frac{{d\ }^2y}{{dx}^2}\ =\frac{1}{x}$

## Numerical Result

The first derivative of given function $y=\ x\ \ln{\ (x)}$ is:

$\frac{dy}{dx}= \ln{(x)}+ 1$

The second derivative of the given function $y=\ x\ \ln{\ (x)}$ is:

$\frac{{d\ }^2y}{{dx}^2}\ =\frac{1}{x}$

## Example

Find out first and second derivative of the function $y=\sqrt x\ \ln{\ (x)}$

Given function:

$y=\sqrt x\ \ln{\ (x)}$

For first derivative, take derivative with respect to $x$ on both sides. We get:

$\frac{dy}{dx}=\frac{d}{dx}\ \left[\sqrt x\ \ln{\ (x)}\right]$

$\frac{dy}{dx}=\frac{d}{dx}[\ \sqrt x\ ] ln(x)+ \sqrt x\frac{d}{dx} [ln(x)]$

$\frac{dy}{dx}=\frac{1}{2\ \sqrt x}\ \ \ln{(x)}+\sqrt x\ \frac{1}{x}\$

$\frac{dy}{dx}=\frac{\ln{(x)}}{2\ \sqrt x}\ +\ \frac{\sqrt x}{x}$

$\frac{dy}{dx}=\frac{\ln{(x)}}{2\ \sqrt x}\ +\ \frac{1}{\sqrt x}$

$\frac{dy}{dx}=\frac{\ln{(x)\ +\ 2}}{2\ \sqrt x}$

To find the second derivative, we will take the derivative of first derivative with respect to $x$ on both sides again.

$\frac{d}{dx}\left(\frac{dy}{dx}\right)\ =\frac{d}{dx}\ \left(\frac{\ln{(x)\ +\ 2}}{2\ \sqrt x}\right)$

$\frac{{d\ }^2y}{{dx}^2}\ =\ \ \frac{2\ \sqrt x\ \frac{d}{dx}(\ln{(x)\ +\ 2)\ -\ (\ln{(x)\ +\ 2)\ \frac{d}{dx}\ \ \left(2\ \sqrt x\right)}}}{\left(2\ \sqrt x\right)^2}$

$\frac{{d\ }^2y}{{dx}^2}\ =\ \ \frac{2\ \sqrt x\ (\ \frac{1}{x}{\ +\ 0)\ -\ (\ln{(x)\ +\ 2)\ \ \left(2\ \times\frac{1}{2\ \sqrt x}\right)}}}{\left(2\ \sqrt x\right)^2}$

$\frac{{d\ }^2y}{{dx}^2}\ =\ \ \frac{2\ \sqrt x\ (\ \frac{1}{x}{\ )\ -\ (\ln{(x)\ +\ 2)\ \ \left(\frac{1}{\ \sqrt x}\right)}}}{\left(2\ \sqrt x\right)^2}$

$\frac{{d\ }^2y}{{dx}^2}\ =\ \ \frac{\ \ \frac{2\ \sqrt x}{x}{\ \ -\ \frac{(\ln{(x)\ +\ 2)\ \ }}{\ \sqrt x}{\ \ }}}{\left(2\ \sqrt x\right)^2}$

$\frac{{d\ }^2y}{{dx}^2}\ =\ \ \frac{\ \ \frac{2\ }{\sqrt x}{\ \ -\ \frac{(\ln{(x)\ +\ 2)\ \ }}{\ \sqrt x}{\ \ }}}{\left(2\ \sqrt x\right)^2}$

$\frac{{d\ }^2y}{{dx}^2}\ =\ \ \frac{\ \ {\ \frac{2\ -\ (\ln{(x)\ +\ 2)\ \ }}{\ \sqrt x}{\ \ }}}{\left(2\ \sqrt x\right)^2}$

$\frac{{d\ }^2y}{{dx}^2}\ =\ \ \frac{\ \ {\ \frac{2\ -\ln{(x)\ -\ 2\ \ }}{\ \sqrt x}{\ \ }}}{\left(2\ \sqrt x\right)^2}$

$\frac{{d\ }^2y}{{dx}^2}\ =\ \ \frac{\ \ {\ \frac{\ -\ln{(x)\ }}{\ \sqrt x}{\ \ }}}{\left(2\ \sqrt x\right)^2}$

$\frac{{d\ }^2y}{{dx}^2}\ =\ \ \frac{\ \ {\ \frac{\ -\ln{(x)\ }}{\ \sqrt x}{\ \ }}}{4x}$

$\frac{{d\ }^2y}{{dx}^2}\ =\ \ \frac{\ -\ln{(x)\ }}{\ 4x\sqrt x}$

The first derivative of the given function $y=\sqrt x\ \ln{\ (x)}$ is:

$\frac{dy}{dx}=\frac{\ln{(x)\ +\ 2}}{2\ \sqrt x}$

The second derivative of the given function $y=\sqrt x\ \ln{\ (x)}$ is:

$\frac{{d\ }^2y}{{dx}^2}\ =\ \ \frac{\ -\ln{(x)\ }}{\ 4x\sqrt x}$