# Strange Derivative

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.

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

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

Last edited:
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.

Last edited:
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?

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
Homework Helper
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.