Question

If $\mathrm{y}={\left(\frac{\mathrm{x}}{\mathrm{n}}\right)}^{\mathrm{nx}}\left(1+\log \left(\frac{\mathrm{x}}{\mathrm{n}}\right)\right)$, then the value of $\mathrm{y}\text{'}\left(\mathrm{n}\right)$ is given by

• A. $\frac{1}{\mathrm{n}}$
• B. ${\left(\frac{1}{\mathrm{n}}\right)}^{\mathrm{n}}$
• C. $\frac{\left[{\mathrm{n}}^2+1\right]}{\mathrm{n}}$
• D. ${\left(\frac{1}{\mathrm{n}}\right)}^{\mathrm{n}}\left(\frac{\left[{\mathrm{n}}^2+1\right]}{\mathrm{n}}\right)$

Answer 1

The correct option is C

$\frac{\left[{\mathrm{n}}^2+1\right]}{\mathrm{n}}$

Find the value of $\mathrm{y}\text{'}\left(\mathrm{n}\right)$:

The given differential equation is $\mathrm{y}={\left(\frac{\mathrm{x}}{\mathrm{n}}\right)}^{\mathrm{nx}}\left(1+\log \left(\frac{\mathrm{x}}{\mathrm{n}}\right)\right)$

Taking $\log$ on both sides,

$$\begin{gathered} \log y={\left(\frac{\mathrm{x}}{\mathrm{n}}\right)}^{\mathrm{nx}}\left(1+\log \left(\frac{\mathrm{x}}{\mathrm{n}}\right)\right) \\ \begin{array}{l}\log y=\left(\mathrm{nx}\right)\log \left(\frac{\mathrm{x}}{\mathrm{n}}\right)+\log \left(1+\log \left(\frac{\mathrm{x}}{\mathrm{n}}\right)\right)\end{array}\mathbf{[}\mathbf{\because }\mathbf{}{\mathbf{loga}}^{\mathbf{b}}\mathbf{=}\mathbf{bloga}\mathbf{]} \end{gathered}$$

Differentiate the above equation with respect to $x$.

$\begin{array}{l}\left(\frac{1}{\mathrm{y}}\right)\mathrm{y}\text{'}=\mathrm{nx}\left(\frac{\mathrm{n}}{\mathrm{x}}\right)\left(\frac{1}{\mathrm{n}}\right)+\log \left(\frac{\mathrm{x}}{\mathrm{n}}\right)\mathrm{n}+\left[\frac{1}{\left(1+\log \left(\frac{\mathrm{x}}{\mathrm{n}}\right)\right)}\left(\frac{\mathrm{n}}{\mathrm{x}}·\frac{1}{\mathrm{n}}\right)\right]\\ \Rightarrow \mathrm{y}\text{'}=\mathrm{y}\left[\mathrm{n}+\left[\frac{1}{\mathrm{x}\left(1+\log \left({\displaystyle \frac{\mathrm{x}}{\mathrm{n}}}\right)\right)}\right]+\log \left(\frac{\mathrm{x}}{\mathrm{n}}\right)\mathrm{n}\right]\end{array}$

Substitute $\mathrm{x}=\mathrm{n}$ in the above equation and in given equation

$\begin{array}{rcl}\mathrm{y}& =& {\left(\frac{n}{\mathrm{n}}\right)}^{nn}\left(1+\log \left(\frac{n}{\mathrm{n}}\right)\right)\\ & =& 1\end{array}$

$\begin{array}{l}\mathrm{y}\text{'}\left(\mathrm{n}\right)=1\left[\mathrm{n}+\left(\frac{1}{\mathrm{n}}\right)\right]\left[\because \log 1=0\right]\\ \Rightarrow \mathrm{y}\text{'}\left(\mathrm{n}\right)=\frac{\left({\mathrm{n}}^2+1\right)}{\mathrm{n}}\end{array}$

Hence, the correct option is C.