Implicit differentiation

  • Thread starter daster
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  • #1
daster
Say we have two functions of x, v and y, such that v=x+y. How can I find v'?
 

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  • #2
dextercioby
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Why doncha differentiate its definition wrt to "x"??

Daniel.

PS.Tell me what u get.
 
  • #3
HallsofIvy
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Since v and y are functions of x, I presume by " v' " you mean "the derivative of v with respect to x", i.e. dv/dx.

Use the chain rule: v= x+ y so dv/dx= dx/dx+ dy/dx= 1+ dy/dx. What dy/dx is depends, of course, on what function y is of x.
 
  • #4
daster
Oh, so it's only dy/dx? Cause I remember my book doing something like (dy/dx)(dv/dx) or something. Thanks HallsofIvy. :smile:

Another question. How do I find d(e^u dy/dx)/du, where u and y are functions of x?
My book says:

[tex]e^u \frac{dy}{dx} + e^u \frac{d^2y}{dx^2} \cdot \frac{dx}{du}[/tex]

I understand this is the product rule, but where'd the dx/du come from?
 
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  • #5
Galileo
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It comes from the chain rule again:

[tex]\frac{d}{du} \frac{dy}{dx}=\frac{d^2y}{dx^2}\frac{dx}{du}[/tex]
 
  • #6
daster
Dank je. :smile:
 
  • #7
Galileo
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Graag gedaan. :biggrin:
 

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