What is the derivative of the inner product norm on $L^2$ space? Let $f \in L^2(X)$ such that $f$ is generated by some arbitrary constant; that is, $f = g(a)$ with $g: \mathbb{R} \to L^2(X)$. Then what can be said about the derivative with respect to some arbitrary variable of the inner product of $f$ with itself:
$$D =\frac{\partial}{\partial a}\langle f, f \rangle = \;?$$
I had the idea that $D = 2 \langle f ,  \frac{\partial f}{\partial a} \rangle$, where $I$ is the identity function, but I'm not sure. The evidence for this is as follows
\begin{equation}
\begin{aligned}
D &=\frac{\partial}{\partial a} \int_X \left(f(x)\right)^2\ dx\\
&= 2\int_X f(x)  \frac{\partial f}{\partial a} \ dx.
\end{aligned}
\end{equation}
My ultimate goal is to compute this for general vector spaces not just $L^2$. Thanks.
 A: Using information from Differentiating an Inner Product, one can approach this problem from a more general perspective.

If $V$, $W$, and $Z$ are normed spaces, and if $T:V\times W\to Z$ is a
  continuous (real) bilinear operator], meaning that there exists
  $C\geq 0$ such that $\|T(v,w)\|\leq C\|v\|\|w\|$ for all $v\in V$ and
  $w\in W$, then the derivative of $T$ at $(v_0,w_0)$ is
  $DT|_{(v_0,w_0)}(v,w)=T(v,w_0)+T(v_0,w)$.  (I am assuming that
  $V\times W$ is given a norm equivalent with
  $\|(v,w)\|=\sqrt{\|v\|^2+\|w\|^2}$.)  This follows from the
  straightforward computation 
$$\frac{\|T(v_0+v,w_0+w)-T(v_0,w_0)-(T(v,w_0)+T(v_0,w))\|}{\|(v,w)\|}=\frac{\|T(v,w)\|}{\|(v,w)\|}\leq C\frac{\|v\|\|w\|}{\|(v,w)\|}\to 0$$
as $(v,w)\to 0$.
With $V=W$, $Z=\mathbb R$ or $Z=\mathbb C$, and $T:V\times V\to Z$ the
  inner product, this gives $DT_{(v_0,w_0)}(v,w)=\langle
 v,w_0\rangle+\langle v_0,w\rangle$.  Now if $f,g:\mathbb R\to V$ are
  differentiable, then $F:\mathbb R\to V\times V$ defined by
  $F(t)=(f(t),g(t))$ is differentiable with $DF|_t(h)=h(f'(t),g'(t))$. 
  By the chain rule, 
$$D(T\circ F)|_{t}(h)
 =DT|_{F(t)}\circ DF|_t(h)=h(\langle f'(t),g(t)\rangle+\langle f(t),g'(t)\rangle),$$
which means $\frac{d}{dt} \langle f, g \rangle = \langle f'(t),g(t)\rangle+\langle f(t),g'(t)\rangle$.

Since Jonas Meyer has shown above the derivative of the inner product, applying this to the $L^2$ inner product norm, we show that
$$\frac{\partial}{\partial a} \langle f, f\rangle = 2\left\langle f , \frac{\partial f}{\partial a} \right\rangle$$ 
A: If $x\rightarrow \dfrac{\partial f}{\partial a}(x,a)\in L^2(X)$ , then $\int_Xf(x,a)\dfrac{\partial f}{\partial a} (x,a)dx$ makes sense. Yet, there is no reason why $D(a)=2\int_Xf(x,a)\dfrac{\partial f}{\partial a} (x,a)dx$. Indeed, this equality is associated to the derivative under the signum $\int_X$; we can do that when, for instance (Lebesgue's theorem) $|f(x,a)\dfrac{\partial f}{\partial a}(x,a)|\leq g(x)$ where $g(x)\in L^1(X)$ ($g$ does not depend on $a$).
If $x\rightarrow \dfrac{\partial f}{\partial a}(x,a)\notin L^2(X)$, then we cannot say anything about $D(a)$.
