Multi Calculus- Partial Derivatives

Homework Statement

I am translating the question from another language so it might not be word to word with the original question.

assume x(s,t) and y(s,t) determined by these two functions:

sin(xt) +x+s=1
eyt+y(s+1)=1

The following function is defined H(x,y)=x2+y2-xy
such that
K(s,t)=H(x(s,t),y(s,t))

find the partial derivative Kt(0,0) (the derivative of K in terms of t)

The Attempt at a Solution

first I wanted to solve the first two equation system in terms of x and y, but it didn't work.
tried also plugging them in to H, no success.

Would appreciate any help.
Roni

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gabbagabbahey
Homework Helper
Gold Member
Start by applying the chain rule to $H$...

Start by applying the chain rule to $H$...
what do you mean by the chain rule ?

should I first replace the x,y with s,t ?
I don't really see the lead ....
would appreciate if you could give any more hints but I'm going to try it also

I tried the chain rule, but looks like I'm doing something wrong.

Kt=Kx*Xt+Ky*Yt

I am assuming K(x,y)=H(x,y), a correct assumption ?
I am getting 0 and that's not the answer.
Kx=2x-y
Ky=2y-x

no matter what I multiply these by, I must get a zero but the answer is not zero.
So I guess I am missing something

gabbagabbahey
Homework Helper
Gold Member
I tried the chain rule, but looks like I'm doing something wrong.

Kt=Kx*Xt+Ky*Yt

I am assuming K(x,y)=H(x,y), a correct assumption ?
Well, sort of. $K(s,t)$ is explicitly a function of $s$ and $t$ only. You know that $x$ and $y$ are also some function of $s$ and $t$ and that

[tex]K(s,t)=H(x(s,t),y(s,t))=[x(s,t)]^2+[y(s,t)]^2+x(s,t)y(s,t)[/itex]

So, it is more correct to say

[tex]K_t(s,t)=H_x x_t+H_y y_t[/itex]

I am getting 0 and that's not the answer.
Kx=2x-y
Ky=2y-x
Well, $H_x=2x-y$ and $H_y=2y-x$....but what about $x_t$ and $y_t$....what are you getting for those?

Well, sort of. $K(s,t)$ is explicitly a function of $s$ and $t$ only. You know that $x$ and $y$ are also some function of $s$ and $t$ and that

[tex]K(s,t)=H(x(s,t),y(s,t))=[x(s,t)]^2+[y(s,t)]^2+x(s,t)y(s,t)[/itex]

So, it is more correct to say

[tex]K_t(s,t)=H_x x_t+H_y y_t[/itex]

Well, $H_x=2x-y$ and $H_y=2y-x$....but what about $x_t$ and $y_t$....what are you getting for those?
Xt=-x*cos(xt)
Yt=(-yeyt)/(t*eyt+s+1)

But it doesnt matter what I'm getting for these because I multiply each one of them by a zero.

ohh hold on, the s and t are zeros , so it doesnt mean x and y are zeros
wow

Edit:

thanks for the help

gabbagabbahey
Homework Helper
Gold Member
Xt=-x*cos(xt)
Yt=(-yeyt)/(t*eyt+s+1)
Your expression for $x_t$ doesn't look quite right.

But it doesnt matter what I'm getting for these because I multiply each one of them by a zero.
I'm not sure why you think this. $H_x=2x-y\neq 0$ and $H_y=2y-x\neq 0$

Your expression for $x_t$ doesn't look quite right.
oh, it didn't influence my answer so I didn't see the mistake

I think it's gotta be
-x*cos(xt)/(t*cos(xt)+1)

Thank you.