Strange Derivative

  • Thread starter Lancelot59
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  • #1
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I'm given these functions:
[tex]f(x,y)=x^{3}y[/tex]
[tex]ye^{y}=t[/tex]
[tex]x^{3}+tx=8[/tex]

I need to find (df/dt)(0)

I have no clue how to go about this. I can't isolate any of the variables. I tried making an implicit function out of equations two and three, but that didn't lead anywhere useful.

I'm stumped.
 

Answers and Replies

  • #2
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I tried making an implicit function out of equations two and three, but that didn't lead anywhere useful.

If you have done this, then try the chain rule (http://en.wikipedia.org/wiki/Chain_rule) but use partial derivatives.

if you have [tex]x(t)[/tex] and [tex]y(t)[/tex] then [tex]\frac{df}{dt}[/tex] can be found by

[tex]\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}[/tex]
 
Last edited:
  • #3
634
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If you have done this, then try the chain rule (http://en.wikipedia.org/wiki/Chain_rule) but use partial derivatives.

How does that help?

[tex]x^{3}+(ye^{y})x=8[/tex]

I don't see how this relates to the original function.

EDIT: Sorry, I posted before you edited. Let me give that a go.

The issue is that I can't get x(t) and y(t). I can only get t(x) and t(y) which isn't too useful.
 
Last edited:
  • #4
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Hmm... you can still get x'(t) and y'(t) by differentiating implicitly t(x) and t(y), but you will be left with derivatives still partially in terms of x and y...

Edit: I am not sure if this is valid, but as you are evaluating the final derivative at t=0, why not do it to all the partial derivatives along the way, hence removing the other variables?
 
  • #5
634
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Hmm... you can still get x'(t) and y'(t) by differentiating implicitly t(x) and t(y), but you will be left with derivatives still partially in terms of x and y...

Edit: I am not sure if this is valid, but as you are evaluating the final derivative at t=0, why not do it to all the partial derivatives along the way, hence removing the other variables?

Sorry, I'm not following along with this.
 
  • #6
hunt_mat
Homework Helper
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This is relatively simple, to find x at t=0, set t=o to fint x^{3}=3=> x=2 at t=0, likewise y=0 at t=0, then differentiate as usual. Not so hard...
 
  • #7
634
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I guess I need more help with derivatives. I'll talk to my prof. Thanks for the help.
 

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