If sin(−2x+2y+z)=0, find the first partial derivatives ∂z∂x and ∂z∂y at the point (0, 0, 0).

Question

anonymous · Answer

shouldn't it be -2cos(-2x+2y+z)?  plug in (0,0,0) and the answer is -2?

ganeshie8 · Answer

yes

anonymous · Answer

@ganeshie8 , that's what i thought, but webwork says the answer is wrong....any idea why?

anonymous · Answer

attachment

ganeshie8 · Answer

oh wait you want to find $\large \dfrac{\partial z}{\partial x}$

anonymous · Answer

yeah, i thought that's what i was doing...i'm lost

ganeshie8 · Answer

sin(−2x+2y+z)=0

cos(−2x+2y+z)*(-2 + $\dfrac{\partial z}{\partial x}$) = 0

ganeshie8 · Answer

plugin (0,0,0) and solve $\dfrac{\partial z}{\partial x}$

anonymous · Answer

what happened to the "2y" in the second part?  -2 is the derivative of 2x, but shouldn't 2 be the derivative of 2y and derivative of z be 1?

anonymous · Answer

@ganeshie8

ganeshie8 · Answer

y is a constant when you're `differentiating z implicitly with respect to x`

anonymous · Answer

so for the second part, it would be (2+dz/dy)?  the x component is constant and differentiate the y?

ganeshie8 · Answer

yes set that equal to 0 and you get dz/dy = -2

anonymous · Answer

ok.  thanks.  got both parts right now.

ganeshie8 · Answer

good, you could also work it by solving z explicitly before differentiating

ganeshie8 · Answer

sin(−2x+2y+z)=0

-2x + 2y + z = arcsin0)

z = arcsin(0) + 2x - 2y

anonymous · Answer

that's a good point.  thanks a bunch.