1. Oct 6, 2015

### BigFlorida

I have a multivariable function z = x2 + 2y2 such that x = rcos(t) and y = rsin(t). I was asked to find (I know the d's should technically be curly, but I am not the best at LaTeX). I thought this would just be a simple application of chain rule:
2/(∂y∂t) = (∂z/∂x)(ⅆx/ⅆt) + (∂z/∂y)(ⅆy/ⅆt)
but apparently this is not the case. Could someone perhaps explain why this is not the right thing to do. When I did it this way I got 2x = 2rcos(t) = 2ycot(t) as my answer, but my book says the answer is -2y2cot(t)csc2(t) and I have no clue what they are doing.

Any help would be appreciated. Thank you in advance.

2. Oct 6, 2015

### SteamKing

Staff Emeritus
The derivative, $\frac {∂^2z}{∂y∂t}$ is known as a mixed partial derivative. Since the expression contains only the function z differentiated w.r.t. y and t, there is no need to differentiate z w.r.t. x. You may differentiate z w.r.t. y first, then w.r.t. t next.

3. Oct 6, 2015

### BigFlorida

@SteamKing But does it not matter that x is a function of t, and (through the relation with r) a function of y? And the only way the book's answer makes sense is if you substituted rcos(t) = ycot(t) into x^2 then differentiate that, treating 2y^2 as a constant, which knocks it out completely.

4. Oct 6, 2015

### Staff: Mentor

I'm confused as to what you're doing. In the first paragraph, you write $\frac{\partial^2 z}{\partial t \partial y}$, but later you write the opposite order, $\frac{\partial^2 z}{\partial y \partial t}$. The first mixed partial is the same as this: $\frac{\partial }{\partial t} \frac{\partial z}{\partial y}$. In the second mixed partial, the differentiation occurs in the opposite order.

Here's the LaTeX I used, tweaked a bit so that it won't render:
$\frac{\partial^2 z}{\partial t \partial y}$
$\frac{\partial^2 z}{\partial y \partial t}$
$\frac{\partial }{\partial t} \frac{\partial z}{\partial y}$