Question

Find \(\dfrac{\partial z}{\partial x},\dfrac{\partial z}{\partial y}\). (you may need tot use implicit differentiation)

\[\sin z^2+e^u =0\qquad\qquad\text{where }u =\tan^{-1}(x-y)\]

Answer 1

\[u_x =\frac1{1+(x-y)^2}\\u_y =-\frac1{1+(x-y)^2}\]

Answer 2

Why? Like, for real, why?

Answer 3

why not?

Answer 4

It's one of those things you'll get eventually so I just want to know if it's a test or you really do need a solution. Because right now, I just want to watch anime.

Answer 5

well, i have the answers,

\[z_x=-\frac{e^u}{2z\cos z^2(1+(x-y)^2)}\\
z_y=\frac{e^u}{2z\cos z^2(1+(x-y)^2)}\]

but i cant work out the proper method

Answer 6

well observe:$$\frac{\partial}{\partial x}[\sin z^2+e^u]=0\\2z\cos(z^2)\frac{\partial z}{\partial x}+e^u\frac{\partial u}{\partial x}=0\\2z\cos(z^2)\frac{\partial z}{\partial x}=-e^u\frac{\partial u}{\partial x}\\\frac{\partial z}{\partial x}=-\frac{e^u}{2z\cos z^2}\frac{\partial u}{\partial x}=-\frac{-e^u}{2z\cos z(1+(x-y)^2)}$$

Answer 7

$$\frac{\partial}{\partial y}[\sin z^2+e^u]=0\\2z\cos(z^2)\frac{\partial z}{\partial y}+e^u\frac{\partial u}{\partial y}=0\\\frac{\partial z}{\partial y}=-\frac{e^u}{2z\cos z^2}\frac{\partial u}{\partial y}=\frac{e^u}{2z\cos z^2(1-(x-y)^2)}$$

Answer 8

then just substitute in \(u=\arctan(x-y)\)... anyways this is basically the same procedure as when we do implicit ordinary differentiation

Answer 9

THANKYOU @oldrin.bataku