How to take a derivative of a function with messy parameters $\frac{d}{dt} f(x\cos(t) + y\sin(t), y\cos(t) - x\cos(t), z)$ How would you compute
$$\frac{d}{dt} f(x\cos(t) + y\sin(t), y\cos(t) - x\cos(t), z)$$
in terms of the original function $f$ and $x,y,t$? Assume $f$ is some well-behaved but not specified function. 
Note: This is a total derivative wrt $t$, not partial.
 A: Well, I'll solve this using the chain rule written in it's most useful manner: in terms of the total derivatives. Let $g: \mathbb{R}^n \to \mathbb{R}^k$ be differentiable at $a \in \mathbb{R}^n$ with derivative $Dg(a)$ and let $f: \mathbb{R}^k \to \mathbb{R}^m $ be differentiable at $g(a) \in \mathbb{R}^k$ with derivative $Df(g(a))$,  then the composition $f \circ g : \mathbb{R}^n \to \mathbb{R}^m$ is differentiable at $a$ and we have $D(f\circ g)(a) = Df(g(a))\circ Dg(a)$. If you've never seen this way of writting the chain rule before, take a look at Spivak's Calculus on Manifolds.
Well, now we have $f:\mathbb{R}^3 \to \mathbb{R}$ and $\alpha: \mathbb{R}\to \mathbb{R}^3$, and I'll assume that $(x,y,z) \in \mathbb{R}^3$ is fixed so that we have: $\alpha(t) = (x\cos t+ y \sin t, y\cos t - x \cos t,z)$. Now, it's pretty clear that we have $f \circ \alpha : \mathbb{R} \to \mathbb{R}$ and we want it's derivative.
Using what I told above, we'll have $D(f\circ \alpha)(t) = Df(\alpha(t)) \circ D\alpha(t)$. Remember, however, that the derivative of a function is a linear transformation, such that in the canonical bases of domain and codomain the matrix of the transformation is composed by the partial derivatives of the function. In terms of matrices then, this composition becomes:
$$(f\circ \alpha)'(t)=f'(\alpha(t))\alpha'(t)$$
Look that $f'(\alpha(t))$ is a row matrix and $\alpha'(t)$ is a column matrix so that the multiplication gives a scalar which is the derivative of a function $\mathbb{R} \to \mathbb{R}$ as expected. Now, we have:
$$\alpha'(t)=\begin{pmatrix}-x\sin t + y \cos t \\ -y \sin t + x \sin t \\ 0\end{pmatrix}$$
And we also have:
$$f'(\alpha(t)) = \begin{pmatrix}D_1f(\alpha(t)) & D_2f(\alpha(t)) & D_3f(\alpha(t))\end{pmatrix}$$
And the multiplication gives the derivative we want:
$$(f\circ\alpha)'(t)=(-x\sin t+ y\cos t)D_1f(\alpha(t))+(-y\sin t + x\sin t)D_2f(\alpha(t))$$
Just to finish a little geometrical interpretation: $\alpha$ is a curve in $3$-space and when you compose $f$ with $\alpha$ you are calculating the restriction of $f$ to the curve. If you imagine that $f$ measures temperatures somewhere, $\alpha$ is a curve on this somewhere and $f\circ \alpha$ is the temperature along the curve. Note that $(f\circ\alpha)'(t)$ is a measure of the rate of change of $f$ along $\alpha$ at $\alpha(t)$ so that in the end you are really calculating the directional derivative of $f$ along a curve.
Now a comment might motivate doing things this way: I've already worked with both ways of writing the chain rule, this way, and the other one that deals directly with the partials. Personally this way suits me better, because I find the other way a little confusing. Of course the other way is derived from this one that is far more general, however, this way makes things really clean and simple. If you never saw this way of working with derivatives (as linear transformations) take a look on Spivak's Calculus on Manifolds and Apostol's Calculus Vol. 2.
