Question

If \[u\left( {x,y} \right) = y\log x + x\log y\], then \[{u_x}{u_y} - {u_x}\log x - {u_y}\log y + \log x\log y = \]
A. \[0\]
B. \[ - 1\]
C. \[1\]
D. \[2\]

Answer 1

Hint: In the given question we need to find the value of \[{u_x}{u_y} - {u_x}\log x - {u_y}\log y + \log x\log y\]. For that, we first differentiate the given expression with respect to x as well as with respect to y by the product rule of differentiation and then substitute the value in the required expression to get the desired result.

Formula used:
We have been using the following formulas:
1. The product rule of differentiation is \[\dfrac{d}{{dx}}\left( {u.v} \right) = v\dfrac{{du}}{{dx}} + u\dfrac{{dv}}{{dx}}\]
2. \[\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}\]

Complete step-by-step answer:
We are given that \[u\left( {x,y} \right) = y\log x + x\log y...\left( 1 \right)\]
We are asked to find the value of \[{u_x}{u_y} - {u_x}\log x - {u_y}\log y + \log x\log y...\left( 2 \right)\]
Now first we differentiate the equation \[\left( 1 \right)\] with respect to \[x\], we get
\[
  {u_x} = y \cdot \dfrac{1}{x} + \log \cdot 0 + \log y.1 + x \cdot 0 \\
   = \dfrac{y}{x} + \log y...\left( 3 \right) \\
 \]
Now first we differentiate the equation \[\left( 1 \right)\] with respect to \[y\], we get
\[
  {u_y} = 1.\log x + y \cdot 0 + x.\dfrac{1}{y} + \log y \cdot 0 \\
   = \log x + \dfrac{x}{y}...\left( 4 \right) \\
 \]
Now we substitute the value from equation \[\left( 3 \right)\] and equation \[\left( 4 \right)\] in equation \[\left( 2 \right)\]:
\[{u_x}{u_y} - {u_x}\log x - {u_y}\log y + \log x\log y = \left( {\dfrac{y}{x} + \log y} \right)\left( {\log x + \dfrac{x}{y}} \right) - \left( {\dfrac{y}{x} + \log y} \right)\log x - \left( {\log x + \dfrac{x}{y}} \right)\log y + \log x\log y\]
Now we simplify the above equation, we get
\[
  {u_x}{u_y} - {u_x}\log x - {u_y}\log y + \log x\log y = \dfrac{y}{x}\log x + \dfrac{y}{x} \times \dfrac{x}{y} + \log y\log x - \dfrac{y}{x}\log x - \log y\log x - \log x\log y \\
   - \dfrac{x}{y}\log y + \log x\log y \\
 \]
On further simplification, we get
\[
  {u_x}{u_y} - {u_x}\log x - {u_y}\log y + \log x\log y = {{\dfrac{y}{x}\log x}} + \dfrac{y}{x} \times \dfrac{x}{y}{{ + \log y\log x}}{{ - \dfrac{y}{x}\log x}}{{ - \log y\log x}}{{ - \log x\log y}} \\
  {{ - \dfrac{x}{y}\log y}} + {{\log x\log y}} \\
   = \dfrac{y}{x} \times \dfrac{x}{y} \\
   = 1 \\
 \]
Therefore, the value of \[{u_x}{u_y} - {u_x}\log x - {u_y}\log y + \log x\log y\] is equal to \[1\] when \[u\left( {x,y} \right) = y\log x + x\log y\]

Hence, option (C) is correct

Additional Information: The formula of product rule is a formula used to calculate the derivatives of two or more functions. Assume \[u\] and \[v\] are differentiable functions. As a result, the product of the functions \[u \cdot v\] is differentiable and can be written as \[\dfrac{d}{{dx}}\left( {u.v} \right) = v\dfrac{{du}}{{dx}} + u\dfrac{{dv}}{{dx}}\]

Note: Students must be very careful when differentiating the given expression with respect to \[x\] and \[y\], as well as when substituting the values after differentiating in the required expression to obtain the desired result, as a simple mistake results in the wrong answer.