Multi Calculus- Partial Derivatives

[Roni1985]

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
e^yt+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 K_t(0,0) (the derivative of K in terms of t)

Homework Equations

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 into H, no success.

Would appreciate any help.
Roni

 

[gabbagabbahey]

Start by applying the chain rule to H...

 

[Roni1985]

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

 

[Roni1985]

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

K_t=K_x*X_t+K_y*Y_t

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

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

K_t=K_x*X_t+K_y*Y_t

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

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]<br /> <br /> So, it is more correct to say <br /> <br /> K_t(s,t)=H_x x_t+H_y y_t[/itex]&lt;br /&gt; &lt;br /&gt; &lt;blockquote data-attributes=&quot;&quot; data-quote=&quot;&quot; data-source=&quot;&quot; class=&quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&quot;&gt; &lt;div class=&quot;bbCodeBlock-content&quot;&gt; &lt;div class=&quot;bbCodeBlock-expandContent js-expandContent &quot;&gt; I am getting 0 and that&amp;#039;s not the answer.&lt;br /&gt; K&lt;sub&gt;x&lt;/sub&gt;=2x-y&lt;br /&gt; K&lt;sub&gt;y&lt;/sub&gt;=2y-x &lt;/div&gt; &lt;/div&gt; &lt;/blockquote&gt;&lt;br /&gt; Well, H_x=2x-y and H_y=2y-x...but what about x_t and y_t...what are you getting for those?

 

[Roni1985]

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

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]<br /> <br /> So, it is more correct to say <br /> <br /> K_t(s,t)=H_x x_t+H_y y_t[/itex]&lt;br /&gt; &lt;br /&gt; &lt;br /&gt; &lt;br /&gt; Well, H_x=2x-y and H_y=2y-x...but what about x_t and y_t...what are you getting for those?

&lt;br /&gt; &lt;br /&gt; X&lt;sub&gt;t&lt;/sub&gt;=-x*cos(xt) &lt;br /&gt; Y&lt;sub&gt;t&lt;/sub&gt;=(-ye&lt;sup&gt;yt&lt;/sup&gt;)/(t*e&lt;sup&gt;yt&lt;/sup&gt;+s+1)&lt;br /&gt; &lt;br /&gt; But it doesn&amp;#039;t matter what I&amp;#039;m getting for these because I multiply each one of them by a zero.

 

[Roni1985]

ohh hold on, the s and t are zeros , so it doesn't mean x and y are zeros
wow

Edit:
got the answer,

thanks for the help

 

[gabbagabbahey]

X_t=-x*cos(xt)
Y_t=(-ye^yt)/(t*e^yt+s+1)

Your expression for x_t doesn't look quite right.

But it doesn't 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

 

[Roni1985]

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 got to be
-x*cos(xt)/(t*cos(xt)+1)

Thank you.