Partial derivatives and chain rule

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  • #1
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Homework Statement



express [tex](\frac{\partial u}{\partial s})_{v}[/tex] in terms of partial derivatives of u(s,t) and t(s,v)

Homework Equations





The Attempt at a Solution



I'm pretty stuck with this problem. I know that

dv = [tex](\frac{\partial v}{\partial s})_{t} ds + (\frac{\partial v}{\partial t})_{s} dt[/tex]

and similarly for u(s,t) and t(s,v). But problem is that where do I get [tex](\frac{\partial u}{\partial s})_{v}[/tex] from, and how to express it?
 

Answers and Replies

  • #2
UD1
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d/(dv)[du/ds]=

d/(dv)[du(s,t(s,v))/ds]=

d/(dv)[du/ds + du/dt * dt/ds]=

d/(dv)[du/ds+du(s,t(s,v))/dt * dt(s,v)/ds]=

d^2u/dsdt * dt/dv + d^2u/dt^2 * dt/dv * dt/ds + du/dt* d^t/dsdv

Write it down in usual partial derivative notation and you'll see where it all comes from.
 
  • #3
gabbagabbahey
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d/(dv)[du/ds]=

d/(dv)[du(s,t(s,v))/ds]=

d/(dv)[du/ds + du/dt * dt/ds]=

d/(dv)[du/ds+du(s,t(s,v))/dt * dt(s,v)/ds]=

d^2u/dsdt * dt/dv + d^2u/dt^2 * dt/dv * dt/ds + du/dt* d^t/dsdv

Write it down in usual partial derivative notation and you'll see where it all comes from.
Please don't give out complete solutions (even if they are grievously incorrect like this one), it violates forum rules and doesn't help the student learn the material.
 
  • #4
gabbagabbahey
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Homework Statement



express [tex](\frac{\partial u}{\partial s})_{v}[/tex] in terms of partial derivatives of u(s,t) and t(s,v)

Homework Equations





The Attempt at a Solution



I'm pretty stuck with this problem. I know that

dv = [tex](\frac{\partial v}{\partial s})_{t} ds + (\frac{\partial v}{\partial t})_{s} dt[/tex]

and similarly for u(s,t) and t(s,v). But problem is that where do I get [tex](\frac{\partial u}{\partial s})_{v}[/tex] from, and how to express it?
For starters, [itex]u(s,t)[/itex] is a function of both [itex]s[/itex] and [itex]t[/itex], so use the chain rule.
 
  • #5
UD1
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Please don't give out complete solutions (even if they are grievously incorrect like this one), it violates forum rules and doesn't help the student learn the material.
I apologize. The solution is correct though.
 
  • #6
gabbagabbahey
Homework Helper
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I apologize. The solution is correct though.
No, it isn't. You've misinterpreted the question/notation.

[tex]\left(\frac{\partial u}{\partial s}\right)_v\neq\frac{\partial}{\partial v}\left(\frac{\partial u}{\partial s}\right)[/tex]

Instead, [itex]\left(\frac{\partial u}{\partial s}\right)_v[/itex] is defined as the derivative of [itex]u[/itex] w.r.t [itex]s[/itex], while [itex]v[/itex] is held constant. It's a specific type of partial derivative.
 

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