If u = (8x² + y²) (log x - log y), then \(\left( {{\rm{x}}\frac{{\partial {\rm{u}}}}{{\partial {\rm{x}}}}\,{\rm{ + }}\,{\rm{y}}\frac{{\partial {\rm{u}}}}{{\partial {\rm{y}}}}} \right)\) equals

Concept:

A function of more than one independent variable can be partially differentiated wrt to each of the independent variables separately.

For example, if f(x, y) = 2x² + 3y².

Partial differential of f(x, y) wrt x is f_x = \(\frac{\partial f}{\partial x}\) = \(\frac{\partial (2x^2 + 3y^2)}{\partial x}\) = 4x (take y as a constant when differentiate)

Partial differential of f(x, y) wrt y is f_y = \(\frac{\partial f}{\partial y}\) = \(\frac{\partial (2x^2 + 3y^2)}{\partial y}\) = 6y (take x as a constant when differentiate)

Calculation:

Given:

The equation is u = (8x² + y²)(log x - log y)

The required partial differentials are calculated below:

u_x =  16x(log x - log y) + (8x² + y²)(\(\frac{1}{x}\)) = 16xlog x - 16xlog y + 8x + \(\frac{y^2}{x}\)

u_y = 2y(log x - log y) + (8x² + y²)(-\(\frac{1}{y}\)) = 2ylog x - 2ylog y - \(\frac{8x^2}{y}\) - y

• The result is calculated below:

x\(\frac{\partial u}{\partial x}\)+y\(\frac{\partial u}{\partial y}\)

= x(16xlog x - 16xlog y + 8x + \(\frac{y^2}{x}\)) + y(2ylog x - 2ylog y - \(\frac{8x^2}{y}\) - y)

= 16x²log x - 16x²log y + 8x² + y² + 2y²log x - 2y²log y - 8x² - y²

= 16x²log x + 2y²log x - 16x²log y - 2y²log y

= 2log x(8x² + y²) - 2log y(8x² + y²)

= 2(8x² + y²)(log x - log y)

= 2u

Hence, the correct answer is option 2.