Question

Partial derivative question

Answer 1

\[\frac{ \delta z }{ \delta x } \] of \[x^2 - (y+z)^2 = 0\]

Answer 2

I got 1 but its somehow a different answer

Answer 3

Can't be that 1 is substituted back for the variables

Answer 4

So...

\[\ f(x,y) = z = x^2 -(y+z)^2=0\]\[\ \frac{ \delta z }{ \delta x } = 2x \]

Answer 5

Do you have to solve for x as well? i'm unsure if you set this equal to 0 to solve for x.

Answer 6

x = y + z

Answer 7

Answer is \[\frac{ x }{ y+z }\] but getting there is not the same?

Answer 8

I solved for Z:
\[x^2 - (y+z)^2 = 0\]
\[-(y+z)^2 = -x^2\] \[(y+z)^2 = x^2\] \[y + z = x\] \[z = x- y\] \[\frac{ \delta z }{ \delta x } = 1\]
where \[1 = \frac{ x }{ y+z }\]

Answer 9

refuse to believe the last part i inferred there

Answer 10

Was already set up at = 0

Answer 11

So, this means we are differentiating z with respect to x while y is held constant. So this is just implicit differentiation with y's as constants. So taking the derivative we would have:
\[2x - 2(y+z)(0+ \frac{ \delta z }{ \delta x }) =0\]
I'm okay to differentiate any x-variables, hence why I get 2x. The 2nd term is chain rule with y as a constant. I placed the 0 in the chain rule to make it more visible what I did. From here we just solve for dz/dx
\[2x - 2(y+z)(0+\frac{ \delta z }{ \delta x }) = 0 \implies 2x = 2(y+z)\frac{ \delta z }{ \delta x } \implies \frac{ 2x }{ 2(y+z) } = \frac{ x }{ y+z } = \frac{ \delta z }{ \delta x }\]

Answer 12

And if x = y+z as you posted above, then you would end up with x/x = dz/dx and dz/dx would equal 1

Answer 13

NICEEE

Answer 14

Thanks guys

Answer 15

No problem :)

Answer 16

So that's how this works :) (I'm learning this as well in class)

Answer 17

Yeah, it's treated as implicit differentiation with the variables not mentioned in the partial treated as constants :P

Answer 18

why did you set 2x = 2x dz/dx?

Answer 19

Well, if I have 2x - 2(y+z)(0+dz/dx) = 0 after differentiating, I'm just simplifying it down and trying to solve for dz/dx

2x - 2(y+z)(0+ dz/dx) = 0
2x - 2(y+z)dz/dx = 0
2x = 2(y+z)dz/dx
2x/2(y+z) = dz/dx
x/(y+z) = dz/dx
All I want to do is somehow isolate dz/dx, so thats all that was done. Seemed easier algebraically to move it as I did.

Answer 20

Exactly I missed the + sign with the Zero term but nice this works out pretty well. Thank you for the detailed explanation, seriously. Have to look at implicit again before I make this mistake again

Answer 21

I'm going to retry this problem just to understand it better, haha