Question

If u = log x^2/y, then xdu/dx+ydu/dy is equal to​

Answer 1

Answer:I am not sure about the answer,still

Step-by-step explanation:

u=log((x^2)/y)

partially differentiating,

du/dx=2/x

du/dy=-1/((x^2)y)

so, xdu/dx+ydu/dy=2-(1/(x^2))

Answer 2

Answer:

x $\frac{du}{dx}$ + y $\frac{du}{dy}$ = 1

Step-by-step explanation:

Given that u = log $\frac{x^{2}}{y}$

We need to find the value of x $\frac{du}{dx}$ + y $\frac{du}{dy}$

Therefore,  $\frac{du}{dx}$  = $\frac{d}{dx}$ (log $\frac{x^{2}}{y}$)

As we are differentiating with respect to x, y will be treated as constant.

=>  $\frac{1}{\frac{x^{2}}{y} }$ * $\frac{1}{y}$ * 2x

=>   $\frac{y}{x^{2}}$ * $\frac{1}{y}$ * 2x

=  $\frac{2}{x}$

Therefore,  $\frac{du}{dy}$  = $\frac{d}{dy}$ (log $\frac{x^{2}}{y}$)

As we are differentiating with respect to y, x will be treated as constant.

=>  $\frac{1}{\frac{x^{2}}{y} }$ * x² * $\frac{-1}{y^{2}}$

=>  $\frac{y}{x^{2}}$ * x² * $\frac{-1}{y^{2}}$

=  $\frac{-1}{y}$

Therefore, x $\frac{du}{dx}$ + y $\frac{du}{dy}$ = x * $\frac{2}{x}$ + y * $\frac{-1}{y}$

                                   = 2 + (-1)

                                   = 2 - 1

                                   = 1

Therefore, x $\frac{du}{dx}$ + y $\frac{du}{dy}$ = 1