need a reminder on the chain rule.

Let $(u,v,w)=\mathbf{x}(x,y)=(x+6y,3xy,x^2-3y^3)$. Compute $D\mathbf{f}(x,y),\text{ and }\partial (u,v)/\partial(x,y)$

Question

need a reminder on the chain rule.

Let $(u,v,w)=\mathbf{x}(x,y)=(x+6y,3xy,x^2-3y^3)$. Compute $D\mathbf{f}(x,y),\text{ and }\partial (u,v)/\partial(x,y)$

richyw · Answer

I get$$D\mathbf{f}=\left[\begin{matrix} 1 & 6 \ 3y & 3x \ 2x & -6y \end{matrix}\right]$$which is apparantly incorrect but I think that's just a typo in the solutions. The solutions are not given for the next three. could someone give me a quick refresher here?

amistre64 · Answer

your taking the derivative of a vector function right?

richyw · Answer

yes

amistre64 · Answer

is (u,v,w) that

r = <u,v,w>

such that:

u = x+6y
v = 3xy
w = x^2 = 3y^3

amistre64 · Answer

$$\frac{dr}{dx}=\frac{\delta r}{\delta u}\frac{du}{dx}+\frac{\delta r}{\delta v}\frac{dv}{dx}+\frac{\delta r}{\delta w}\frac{dw}{dx}$$

amistre64 · Answer

not real sure aboutyour notation tho

richyw · Answer

hmm that looks like the chain rule alright.

richyw · Answer

ok, what I'm actually going to do is post the example question from the textbook. because it is the same as this but It gives the answer. I don't understand what is going on but at least I know the answer so it might be easier for someone to explain to me what's going on!

amistre64 · Answer

that would prolly be best at the moment :)