The derivative of $z=x^2+xy+ y^2$

Question

I have got confused about this problem, what I have thought was differentiating this with respect to $x$ gives -

$\frac{dz}{dx} = 2x + x \frac{dy}{dx} + y + 2y \frac{dy}{dx}$

But, I came across an online video which did it in this way-

$\frac{dz}{dx} = 2x + y$

How is this possible? What am I missing?

This is a three dimensional equation, so $y$ is not a constant, so this person cant treat $y$ as a constant I believe.

You're conflating "implicit derivatives" (your understanding) and "partial derivatives" (the video you saw). — Andrew D. Hwang, Aug 30 '14 at 12:51
Oh ok , so partial derivation is a special case of implicit derivate it seems. — M.S.E, Aug 30 '14 at 12:53
for partial derivatives should one be using $\frac{\partial z}{\partial x}$ rather than $\frac {dz}{dx}$? — Hypergeometricx, Aug 30 '14 at 14:13

score 4 · Accepted Answer · answered Aug 30 '14 at 12:52

4

We have

$$\frac{dz}{dx}=2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}$$ but if $y$ doesn't depend on $x$ then $\frac{dy}{dx}=0$ and then $$\frac{dz}{dx}=2x+y$$

answered Aug 30 '14 at 12:52

Oh ok , so partial derivation is a special case of implicit derivate it seems. Thank you , i was actually really disturbed watching the vedio. Will check this special case. – M.S.E Aug 30 '14 at 12:53

score 0 · Answer 2 · answered Aug 30 '14 at 12:55

0

The issue is simply whether $x$ and $y$ are independent variables. If they are then, the approach in the video is legitimate, although one would normally write it as the partial derivative of $z$ wrt $x$. If $y$ depends on $x$, then your way is correct.

answered Aug 30 '14 at 12:55

almagest

18,380

Got it :) Thanks – M.S.E Aug 30 '14 at 12:56

The derivative of $z=x^2+xy+ y^2$

2 Answers2