How does one show that $f(x,y)=\frac{\langle x,y \rangle}{|x||y|}$ is in $C^1(\mathbb{R^2 \setminus {(0,0))}}$ I want to show that $f(x,y)=\frac{\langle x,y \rangle}{|x||y|}$ is in $C^1(\mathbb{R^2 \setminus \{(0,0)\})}$. 
I have already shown that $\langle x,y \rangle$ is continuous, and thus $f$ is continuous because its a composition of continous functions.
Now I need to find the partial derivatives of $f$ and I have no idea how to differentiate this function.
 A: The map $x \mapsto \langle x, y \rangle$ is a linear function of $x$ so it is its own derivative. The function $x \mapsto \frac{1}{|x|} = \frac{1}{\sqrt{\langle x, x \rangle}}$ is a composition of differentiable functions. So to differentiate we use the chain rule.
First note that the derivative of $x \mapsto \langle x, x\rangle$ is the map $h \mapsto 2\langle x, h \rangle$ since
$$ \langle x + h, x + h \rangle - \langle x, x \rangle = 2\langle x, h \rangle + \langle h, h \rangle = 2\langle x, h \rangle + o(|h|^2) $$
Thus applying the chain rule, the derivative of $x \mapsto \frac{\langle x, y \rangle}{|x||y|}$ is the map
$$ h \mapsto \frac{\langle h, y \rangle}{|x||y|} - \frac{\langle x, y \rangle}{2|x|^{3/2}|y|} \cdot 2 \langle x,h \rangle. $$
Or in more traditional notation
$$ \left. \frac{\partial f}{\partial x} \right|_x = \frac{1}{|x||y|}y^T -\frac{\langle x, y \rangle}{|x|^{3/2}|y|} x^T.  $$

Notice that this is complicated. We have theorems that simplify this:


*

*If $f$ is linear then $f$ is smooth

*If $f = gh$ and both $g, h$ are differentiable then so is $f$

*If $f = g/h$, both $g$ and $h$ are differentiable at $x$ and $h \ne 0$ anywhere in a neighbourhood of $x$ then $f$ is differentiable at $x$

*If $f = g \circ h$ and both $g$ and $h$ are differentiable then so is $f$


So to do this without all that mess above, note that $f(x,y) = \frac{\langle x, y \rangle}{|x||y|}$ is the quotient of $\langle x, y \rangle$ and $|x||y|$. The inner product is bilinear and hence differentiable. The norm is $|x| = \sqrt{\langle x, x \rangle}$ is a composition of the smooth functions $t \mapsto \sqrt t$ with $(x, y) \mapsto \langle x, y \rangle$ with $x \mapsto (x,x)$.
