I'm trying to differentiate the function $$z = x^3-y^3$$ where $$ x = \frac{1}{1+t}, \ \ \ \ y = \frac{t}{t+1}$$
I remember there being a proper way to do this using partial differentiation, but I decided to take a stab at it by expressing $x$ and $y$ in terms of $t$ and differentiating it explicitly. Prepare for some messy maths. Hopefully I didn't make any mistakes here, but the point is whether my approach is a valid one:
$$\frac{dz}{dt} = \frac{d}{dt} [(\frac{1}{1+t})^3] - \frac{d}{dt}[ (\frac{t}{t+1})^3]$$
$$\frac{dz}{dt} = [3(\frac{1}{1+t})^2 * \frac{d}{dt}[(1+t)^{-1}]] - [3(\frac{t}{t+1})^2*\frac{d}{dt}[\frac{t}{t+1}]]$$
$$\frac{dz}{dt} = [-3(\frac{1}{1+t})^2*(\frac{1}{1+t})^{2}] - [3 (\frac{t}{t+1})^2 * (\frac{1}{t+1})^2]$$
Erm.. hopefully this isn't too hideous but this is what I have. Is this approach invalid or a poor idea, and if so, why?
Your approach is valid, but it might help to use the chain rule. A shorter version of your answer is$$ \frac{dz}{dt}=-\frac{3(1+t^2)}{(1+t)^4}. $$