+0  
 
0
691
3
avatar

not really sure how to do this,

partially differentiate, with respect to x and y individually,

 

F = 4xy/(x^2+y^2)

 Feb 28, 2016

Best Answer 

 #2
avatar+26387 
+5

partially differentiate \(F = \frac{4xy}{x^2+y^2}\)

 

\(F_x = F \cdot ( \frac{1}{x} - \frac{2x}{x^2+y^2} ) \\ F_y = F \cdot ( \frac{1}{y} - \frac{2y}{x^2+y^2} )\)

 

 

 

laugh

 Feb 28, 2016
 #1
avatar+25 
+5

Not to different to normal differentiation, just have to treat the variable that your not differentiating as a constant.

 

\(\frac{\delta F}{\delta x} = \frac{4y(y^2-x^2)}{(x^2+y^2)^2}\)

and

 

\(\frac{\delta F}{\delta y} = \frac{4x(x^2-y^2)}{(x^2+y^2)^2}\)

 

through the use of quotient rule.

 Feb 28, 2016
edited by BillyK  Feb 28, 2016
 #2
avatar+26387 
+5
Best Answer

partially differentiate \(F = \frac{4xy}{x^2+y^2}\)

 

\(F_x = F \cdot ( \frac{1}{x} - \frac{2x}{x^2+y^2} ) \\ F_y = F \cdot ( \frac{1}{y} - \frac{2y}{x^2+y^2} )\)

 

 

 

laugh

heureka Feb 28, 2016
 #3
avatar
+5

Possible derivation:
d/dx((4 x y)/(x^2+y^2))
Factor out constants:
  =  4 y (d/dx(x/(x^2+y^2)))
Use the quotient rule, d/dx(u/v) = (v ( du)/( dx)-u ( dv)/( dx))/v^2, where u = x and v = x^2+y^2:
  =  ((x^2+y^2) (d/dx(x))-x (d/dx(x^2+y^2)))/(x^2+y^2)^2 4 y
The derivative of x is 1:
  =  (4 y (-(x (d/dx(x^2+y^2)))+1 (x^2+y^2)))/(x^2+y^2)^2
Simplify the expression:
  =  (4 y (x^2+y^2-x (d/dx(x^2+y^2))))/(x^2+y^2)^2
Differentiate the sum term by term:
  =  (4 y (x^2+y^2-d/dx(x^2)+d/dx(y^2) x))/(x^2+y^2)^2
Use the power rule, d/dx(x^n) = n x^(n-1), where n = 2: d/dx(x^2) = 2 x:
  =  (4 y (x^2+y^2-x (d/dx(y^2)+2 x)))/(x^2+y^2)^2
The derivative of y^2 is zero:
  =  (4 y (x^2+y^2-x (2 x+0)))/(x^2+y^2)^2
Simplify the expression:
Answer: |  =  (4 y (-x^2+y^2))/(x^2+y^2)^2

 

 

Possible derivation:
d/dy((4 x y)/(x^2+y^2))
Factor out constants:
  =  4 x (d/dy(y/(x^2+y^2)))
Use the quotient rule, d/dy(u/v) = (v ( du)/( dy)-u ( dv)/( dy))/v^2, where u = y and v = x^2+y^2:
  =  ((x^2+y^2) d/dy(y)-y d/dy(x^2+y^2))/(x^2+y^2)^2 4 x
The derivative of y is 1:
  =  (4 x (-(y (d/dy(x^2+y^2)))+1 (x^2+y^2)))/(x^2+y^2)^2
Simplify the expression:
  =  (4 x (x^2+y^2-y (d/dy(x^2+y^2))))/(x^2+y^2)^2
Differentiate the sum term by term:
  =  (4 x (x^2+y^2-d/dy(x^2)+d/dy(y^2) y))/(x^2+y^2)^2
The derivative of x^2 is zero:
  =  (4 x (x^2+y^2-y (d/dy(y^2)+0)))/(x^2+y^2)^2
Simplify the expression:
  =  (4 x (x^2+y^2-y (d/dy(y^2))))/(x^2+y^2)^2
Use the power rule, d/dy(y^n) = n y^(n-1), where n = 2: d/dy(y^2) = 2 y:
  =  (4 x (x^2+y^2-2 y y))/(x^2+y^2)^2
Simplify the expression:
Answer: |  =  (4 x (x^2-y^2))/(x^2+y^2)^2

 Feb 28, 2016

5 Online Users

avatar
avatar