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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.

Expert Answer

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:

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

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