Question

(UV)
If u = x² + y², v = 2xy, then
a(x,y)
a)4(x² - y²) b)4(x2 + y2) c)2(x2 - y2) d)2(x2 + y2)​

Answer 1

Answer:

(d)

Step-by-step explanation:

d) ( xt+y+z)42=x%2+y*2+Z42t+2t+2xyt2zxt+2zy

=x42+y2+Z242+2(xytyzt+zx)

Hope it helps you

Answer 2

$\displaystyle \sf{ \frac{ \partial (u,v)}{ \partial (x,y)} } = 4( {x}^{2} + {y}^{2} )$

Given :

u = x² - y² and v = 2xy

To find :

The value of ∂(u,v)/∂(x,y) is

(A) 4(x² + y²)

(B) - 4(x² + y²)

(C) 4(x² - y²)

(D) 0

Solution :

Step 1 of 2 :

Write down the functions

Here the given functions are

u = x² - y² and v = 2xy

Step 2 of 2 :

Find the value of ∂(u,v)/∂(x,y)

$\displaystyle \sf{ \frac{ \partial (u,v)}{ \partial (x,y)} }$

$\displaystyle = \begin {vmatrix} \frac{ \partial \sf u}{ \partial \sf x} & \frac{ \partial \sf v}{ \partial \sf x} \\ \\ \frac{ \partial \sf u}{ \partial \sf y} & \frac{ \partial \sf v}{ \partial \sf y} \end{vmatrix}$

$\displaystyle = \begin {vmatrix} \sf \:2x & \sf \:2y \\ \\ \sf \: - 2y & \sf \:2x \end{vmatrix}$

$\displaystyle \sf{ = 4 {x}^{2} + 4 {y}^{2} }$

$\displaystyle \sf{ = 4( {x}^{2} + {y}^{2} )}$

Hence the correct option is (A) 4(x² + y²)

Correct question :

If u = x² - y² and v = 2xy then the value of ∂(u,v)/∂(x,y) is

(A) 4(x² + y²) (B) - 4(x² + y²) (C) 4(x2 - y²) (D) 0