Question

∂z/∂x when 3xz^2 + ye^−z = 0

I got this to answer -3z^2/6xz-ye^-z but a friend of mine didn't get -3z^2 in the denominator, but 3z^2 instead. Can anybody spot where I have acquired this negative?

Answer 1

$$
\cfrac{\partial}{\partial x}\left ( 3xz^2 + ye^{−z} \right )= 3z^2+3x2z\cfrac{\partial z}{\partial x}+\cfrac{\partial y}{\partial x}e^{-z}+y(-1)e^{-z}\cfrac{\partial z}{\partial x}=0\\
\cfrac{\partial z}{\partial x}\left( 6xz-ye^{-z}\right )=-\cfrac{\partial y}{\partial x}e^{-z}-3z^2\\
\cfrac{\partial z}{\partial x}=\cfrac{-\cfrac{\partial y}{\partial x}e^{-z}-3z^2}{6xz-ye^{-z}}\\
\cfrac{\partial z}{\partial x}=\cfrac{\cfrac{\partial y}{\partial x}e^{-z}+3z^2}{ye^{-z}-6xz}\\
$$

Answer 2

First of all, thanks a lot for taking the time to reply with such a detailed response.

My apologies I can't seem to follow it completely.

It was my understanding that one was to treat y as a constant when calculating ∂z/∂x in this instance.

Hence, I found ∂z/∂x = (3x * 2z(∂z/∂x) + z^2 *3) + ye^-z * -1(∂z/∂x) = 0
-> 6xz(∂z/∂x) + 3z^2 - ye^-z(∂z/∂x) = 0
-> ∂z/∂x(6xz - ye^-z) = -3z^2
-> ∂z/∂x = -3x^2/6xz-ye^-z

Answer 3

Am I completely wrong in my approach there? Apologies for not writing that in equation form - I hope you can follow!

Answer 4

Unless stated, you shouldn't treat any variables as constants. Therefore, you should also treat y as a function of x. If it were not a function of x, then it would automatically be zero for that case and the solution would still be valid.

Answer 5

Everything else in your solution looks correct.

Answer 6

That's awesome - thanks so much for going through all this with me. Your time is genuinely appreciated.