Multivariable Calculus - Calculating Derivative Matrix I'm working with Munkres' Analysis on Manifolds.
From chapter 2 (this isn't a homework question):
Given $f: \mathbb R^2 \rightarrow \mathbb R^2 : f(r,\theta)=(r\cos(\theta),r\sin(\theta))$, calculate $Df$.
I can use directional derivatives (evaluated via a limit), but I'm basically at a loss for how to start. A nudge in the right direction would be great.
 A: Here is a somewhat long explanation that might help you understand where the matrix comes from. Given a function $f: M \to N$, the derivative at a point $p$ is a map from the tangent space of $M$ at $p$ to the tangent space of $N$ at $f(p)$. In other words:
$$(Df)_p = T_p M \to T_{f(p)} N.$$
What is $T_p M$? Informally speaking, it is the space of all vectors based at the point $p$. (Analogously for $T_{f(p)} N$.)
How does $Df_p$ map a tangent vector in $T_p M$ to a tangent vector in $T_{f(p)} N$?
What you can do is you can identify a vector $V$ with a curve that goes through the point $p$. That is, you have some curve $\gamma : (-\varepsilon,\varepsilon) \to M$ with the properties:
$\gamma(0) = p$ and $\gamma'(0) = V$.
Since $\gamma$ can be thought of as a set of points in $M$, it makes sense to take the composition $f\circ \gamma: (-\varepsilon,\varepsilon)\to N$, and this gives you a path inside of $N$.
$f\circ\gamma(0) = f(\gamma(0)) = f(p)$ and by definition, $\dfrac{d}{dt}_{t=0} f\circ\gamma(t) = Df_p V$.
Now how does this make sense for your example?
Imagine that you have a path $\gamma(t) = (r(t),\theta(t))$, with $\gamma(0)=(r_0,\theta_0)$ and $\gamma'(0) = (\dot{r}_0,\dot{\theta}_0)$. Then,
$f(\gamma(t)) = (r(t)\cos(\theta(t)), r(t)\sin(\theta(t))$. Taking the derivative at $t=0$ gives you:
$\dfrac{d}{dt}_{t=0}f(\gamma(t)) = (\dot{r}_0\cos(\theta_0) - r_0\sin(\theta_0)\dot{\theta}_0, \dot{r}_0\sin(\theta_0)+r_0\cos(\theta_0)\dot{\theta}_0) = \left[\begin{matrix} \cos(\theta_0) & -r_0\sin(\theta_0) \\ \sin(\theta_0) & r_0\cos(\theta_0)\end{matrix}\right] \left[\begin{matrix} \dot{r}_0 \\ \dot{\theta}_{0}\end{matrix}\right]$
And we see that we get the same exact matrix janmarqz has posted.
A: The derivative happens to be:
$$Df=\left(\begin{array}{cc}
\cos(\theta)&-r\sin(\theta)\\
\sin(\theta)&r\cos(\theta)
\end{array}\right),$$
and as you can see, in the rows are the gradients of the two functions components of $f$.
