# 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

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.