partial derivatives
f(x,y)= x / (x+y)^2

Question

partial derivatives 
f(x,y)= x / (x+y)^2

anonymous · Answer

hi

anonymous · Answer

can you solve for fx and fy ? i already have my own answer. i just want to compare with your answer

psymon · Answer

Oh, okay, so not much explanation needed then. Alright, lemme check.

psymon · Answer

what answers did you get?

anonymous · Answer

i got [ (x+y)^2 - 2x(x+y) ] / (x+y)^4      for fx

anonymous · Answer

0 for fy

psymon · Answer

fx is fine, fy not so much.

anonymous · Answer

hm.. i used quotient rule for fy. i thought fy was 0 since there was no 'y' on the numerator.......

anonymous · Answer

( everything is constant except for y? )

psymon · Answer

Nah, doesnt mean that at all. In fact, I wouldnt do quotient rule. If x is just a constant, I can do x(x+y)^-2 
Which then I would just do chain rule  =     -2x(x+y)^-3  The inner derivative just becomes 1, so all I really had to do is power rule.

anonymous · Answer

so it is -2x(x+y)^3 (1) = -2x(x+y)^-3 should be answer ?

psymon · Answer

Now if I did quotient rule, just to show ya:

$$\frac{ 0(x+y)^{2}-x(2)(x+y) }{ (x+y)^{4} }$$

psymon · Answer

And yep. It would come out the same way with the quotient rule, but with a slight bit more work.

anonymous · Answer

should've thought about changing the equation form.

psymon · Answer

Yeah. For me, I sometimes might literally replace the x's or y's with constants. Like, say something like "let x = 1" Then just make sure I dont lose the 1 and replace it with x in the end. Helps out a lot visually so youre not confusing yourself.

.sam. · Answer

@Psymon  $$\frac{\partial \text{}}{\partial y}\left((x+y)^2\right)=2 (x+y)$$

psymon · Answer

Its coming up as an error for me right now unfortunately. Although if I did something wrong then im not surprised. I screw up more than would be wanted for someone trying to help out x_x

anonymous · Answer

thank you so much^^ im going to post up another partial derivative problem