# Chain rule for implicit differentiation

1. Jan 21, 2012

### Tuomo

I have derivative dx/dt = y(u(t)) * z(u(t)) + u(t)
Now, what is dx/du ? I know the chain rule should help, but I am stuck :-(

2. Jan 22, 2012

### fluidistic

Maybe $\frac{dx}{du}=\frac{dx}{dt} \cdot \frac{dt}{du}$.

3. Jan 23, 2012

### Tuomo

Thanks fluidistics.

Ok, so, what I get is:

$\frac{dx}{du}=\frac{dx}{dt} \cdot \frac{dt}{du}=\left[y(u(t)) \cdot z(u(t)) + u(t)\right] \cdot \frac{dt}{du(t)}$

Now, what the heck is this $\frac{dt}{du(t)}$ ? Any idea how should I treat it?
Is it just a reciprocal of $\frac{du(t)}{dt}$? In other words, can I just divide the original function with the time derivative of u(t)?

$\frac{dx}{du}=\frac{dx}{dt} \cdot \frac{dt}{du}=\left[y(u(t)) \cdot z(u(t)) + u(t)\right] \cdot \left[\frac{du(t)}{dt}\right]^{-1}$

Thanks for help and comments! Its is almost... uhhhh... 15 years since I studied this stuff in university.