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Function composition

  • Thread starter Ikastun
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  • #1
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Homework Statement



Be z=F(u,v,w), u=f(x,y), v=e-αx, w=ln y, get the expression [itex]\partial[/itex]z/[itex]\partial[/itex]x, [itex]\partial[/itex]z/[itex]\partial[/itex]y.

Homework Equations



Chain rule.

The Attempt at a Solution



[itex]\partial[/itex]z/[itex]\partial[/itex]x=[itex]\partial[/itex]z/[itex]\partial[/itex]v*dv/dx=-α e-αx

[itex]\partial[/itex]z/[itex]\partial[/itex]y=[itex]\partial[/itex]z/[itex]\partial[/itex]w*dw/dy=1/y
 

Answers and Replies

  • #2
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When calculating [itex] \frac{\partial z}{\partial x} [/itex],you should consider all variables that depend on x not only v!
 
  • #3
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Thank you. What about the following expression: ∂z/∂x=∂z/∂u*∂u/∂x+∂z/∂v*dv/dx?. Same to ∂z/∂y with w.
 
  • #4
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Thank you. What about the following expression: ∂z/∂x=∂z/∂u*∂u/∂x+∂z/∂v*dv/dx?.
Looks good. You even picked up on the fact that dv/dx is a regular (not partial) derivative.
Same to ∂z/∂y with w.
What did you get?
 
  • #5
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∂z/∂y=∂z/∂u*∂u/∂y+∂z/∂w*dw/dy.
 
  • #6
33,632
5,287
Looks good!
 

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