Question

If $\mathrm{u}(\mathrm{x},\mathrm{y})=\mathrm{ylogx}+\mathrm{xlogy}$ then ${\mathrm{u}}_{\mathrm{x}}{\mathrm{u}}_{\mathrm{y}}-{\mathrm{u}}_{\mathrm{x}}\mathrm{logx}-{\mathrm{u}}_{\mathrm{y}}\mathrm{logy}+\log \mathrm{xlog}\mathrm{y}=$

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

The correct option is C

$1$

Explanation for the correct option:

Finding the value of the expression:

$\mathrm{u}(\mathrm{x},\mathrm{y})=\mathrm{ylogx}+\mathrm{xlogy}$

Differentiating the above equation with respect to $\mathrm{x}$.

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{ylogx}+\mathrm{xlogy}) \\ {\mathrm{u}}_{\mathrm{x}}=\frac{\mathrm{y}}{\mathrm{x}}+\mathrm{logy}\mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}\left(\mathrm{logx}\right)}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{]} \end{gathered}$$

Similarly, differentiating with respect to $\mathrm{y}$.

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dy}}\left(\mathrm{ylog}\mathrm{x}+\mathrm{xlogy}\right) \\ {\mathrm{u}}_{\mathrm{y}}=\log \mathrm{y}+\frac{\mathrm{x}}{\mathrm{y}}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}\mathbf{\left(}\mathbf{logy}\mathbf{\right)}}{\mathbf{dy}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{y}}\mathbf{]} \end{gathered}$$

Substitute these values in the given expression, we get

$\begin{array}{rcl}{\mathrm{u}}_{\mathrm{x}}{\mathrm{u}}_{\mathrm{y}}-{\mathrm{u}}_{\mathrm{x}}\log \mathrm{x}-{\mathrm{u}}_{\mathrm{y}}\mathrm{logy}+\mathrm{logxlog}\mathrm{y}& =& \left[\left(\mathrm{y}/\mathrm{x}\right)+\mathrm{logy}\right]\left[\mathrm{logx}+\left(\mathrm{x}/\mathrm{y}\right)]–[\left(\mathrm{y}/\mathrm{x}\right)+\log \mathrm{y}\right]\mathrm{logx}–[\mathrm{logx}+\left(\mathrm{x}/\mathrm{y}\right)]\mathrm{logy}+\mathrm{logxlogy}\\ & =& \left(\mathrm{y}/\mathrm{x}\right)\mathrm{logx}+1+\mathrm{logylogx}+\left(\mathrm{x}/\mathrm{y}\right)\mathrm{logy}–\left(\mathrm{y}/\mathrm{x}\right)\mathrm{logx}–\mathrm{logylogx}–\mathrm{logxlogy}–\left(\mathrm{x}/\mathrm{y}\right)\mathrm{logy}+\mathrm{logxlogy}\\ & =& 1\end{array}$

Hence, the correct option is C.