Calculation of a derivative I have to calculate the following derivative
$$\frac{\partial}{\partial{\Vert x\Vert}}e^{ix\cdot y}$$
Then I write
$$e^{ix\cdot y}=e^{i\Vert x\Vert\Vert y\Vert\cos\alpha}$$
andI derive; is this reasoning correct?
 A: We have  $\alpha=\alpha(x,y)=\cos^{-1}\left( \left\langle \frac{1}{\|x\|}\cdot x\,,\, \frac{1}{\|y\|}\cdot y\right\rangle \right) $!
\begin{align}
\frac{\partial}{\partial{\Vert x\Vert}}e^{i\langle x, y\rangle}
=
&
\frac{\partial}{\partial{\Vert x\Vert}}e^{\left({\|x\|}\cdot i\left\langle \frac{1}{\|x\|}\cdot x\,,\, y\right\rangle\right)}
\\
=
&
e^{{\|x\|}\cdot i\left\langle \frac{1}{\|x\|}\cdot x\,,\, y\right\rangle}
\frac{\partial}{\partial{\Vert x\Vert}}
\left({{\|x\|}\cdot i\left\langle \frac{1}{\|x\|}\cdot x\,,\, y\right\rangle}\right)
\\
=
&
e^{{\|x\|}\cdot i\left\langle \frac{1}{\|x\|}\cdot x\,,\, y\right\rangle}
\left(
{ i\left\langle \frac{1}{\|x\|}\cdot x\,,\, y\right\rangle}
+
{{\|x\|}\cdot i\left\langle \frac{-1}{\|x\|^2}\cdot x\,,\, y\right\rangle}
\right)
\\
=
&
i\frac{1}{\|x\|}e^{ i\left\langle x\,,\, y\right\rangle}
\left(
{ \left\langle  x\,,\, y\right\rangle}
-
{{\|x\|}\cdot \left\langle x\,,\, y\right\rangle}
\right)
\\
\end{align}
A: I don't know if it was E. Costa's purpose but the answer should be: expression $\frac{\partial}{\partial{\Vert x\Vert}}e^{ix\cdot y}$ is ambiguous and then meaningless.
You can find several different answers depending on how you interprete it. And contrary to what suggest nbubis, $||y||$ and $α$ are not necessarily independent from $||x||$. Dependency is a question of choice. I can rewrite $x\cdot y$ in function of a different set of real-valued parameters and declare them independent.
If you rewrite your expression $\left(\frac{\partial}{\partial{\Vert x\Vert}}e^{ix\cdot y}\right)_{||y||, \alpha}$ using thermodynamicians convention (here subscript "$||y||, \alpha$" means holding $||y||$ and $\alpha$ constants), then the result is actually $i(\Vert y\Vert\cos\alpha )e^{ix\cdot y}$
A: Yes - Since $||\vec{y}||$ and $\alpha$ are not dependent on $||\vec{x}||$.
