Solving Strange Derivative: Find (df/dt)(0)

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I'm given these functions:
f(x,y)=x^{3}y
ye^{y}=t
x^{3}+tx=8

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.
 
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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 x(t) and y(t) then \frac{df}{dt} can be found by

\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}
 
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adoado said:
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?

x^{3}+(ye^{y})x=8

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.
 
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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?
 
adoado said:
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.
 
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...
 
I guess I need more help with derivatives. I'll talk to my prof. Thanks for the help.
 
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