Partial derivative chain rule proof

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The discussion revolves around proving the equation d²u/dx² + d²u/dy² = e^(-2s)[d²u/ds² + d²u/dt²] for the function u = f(x, y) where x = e^s cos(t) and y = e^s sin(t). Participants express confusion about the meaning of the derivative d/ds and its application in the context of the problem. One user notes that their calculations yield six terms, which do not match the expected results, leading to frustration. There is a request for clarification on the problem and the derivative's role in the proof. The conversation highlights the challenges of understanding partial derivatives in relation to the chain rule.
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Homework Statement



If u=f(x,y) where x=escost and y=essint

show that d2u/dx2+d2u/dy2 = e-2s[d2u/ds2+d2u/dt2

Homework Equations



http://s11.postimage.org/sjwt1wkvl/Untitled.jpg

The Attempt at a Solution



ok i don't understand how they got to that


i don't know what d/ds is supposed to do exactly, i thought it was the second partial derivative of s, but it seems that it's not if we check the answer.


i basically got e^s cost (d^2/dx^2) + e^s sint (d^2u/dy^2), but they got 6 terms. i am kinda lost, can you explain to me what d/ds is supposed to do?
 
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Maybe you should try typing the problem a little clearer. I think I could help you with this.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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