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

[Lancelot59]

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.

 

[adoado]

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

$$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}$$

 

[Lancelot59]

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.

 

[adoado]

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?

 

[Lancelot59]

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.

 

[hunt_mat]

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...

 

[Lancelot59]

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