The differential of a definite integral I have come across this as a fundamental theorem of calculus:
$\frac{d}{dx}\int_{a}^{x} f(t) dt = f(x)\tag{1}$
for any constant $a$.
For example here:
http://mathmistakes.info/facts/CalculusFacts/learn/doi/doi.html
I find myself wanting a solution to this however:
$\frac{d}{dx}\int_{-x}^{x} f(t) dt$
and am not finding this situation covered anywhere alas. I could of course easily rewrite this as:
$\frac{d}{dx}\int_{-x}^{0} f(t) dt + \frac{d}{dx}\int_{0}^{x} f(t) dt$
and this oft encountered fundamental theorem of calculus reduces this to:
$\frac{d}{dx}\int_{-x}^{0} f(t) dt + f(x)$
And if I knew $f(x)$ was symmetrical about 0, of course this would easily reduce to $2f(x)$ but I don't know that, in fact I know it's not. What then? Are there any further rules or theorems of calculus that I've forgotten that might help here?
UPDATE: 
Now I'm truly bamboozled and pulling my hair out. Somewhere I am making a fundamentally simple error it seems but I cannot after review, review, and review find where. To wit here is the problem:
From basic integral conventions we know:
$\begin{align}
 \int_a^c f(x) \, dx &{}= \int_a^b f(x) \, dx - \int_c^b f(x) \, dx \tag{2}\\
 &{} = \int_a^b f(x) \, dx + \int_b^c f(x) \, dx\tag{3}
\end{align}$
and:
$\int_a^b f(x) \, dx = - \int_b^a f(x) \, dx\tag{4}$
Applying these we can write:
$\begin{align*}
\frac{d}{dx}\int_{-x}^{x} f(t)\;dt
&= 
\frac{d}{dx}\int_{-x}^{0} f(t)\;dt
+
\frac{d}{dx}\int_{0}^{x} f(t)\;dt
\\
&= 
-\frac{d}{dx}\int_{0}^{-x} f(t)\;dt
+
\frac{d}{dx}\int_{0}^{x} f(t)\;dt
\\
&= 
\frac{d}{dx}\int_{0}^{x} f(t)\;dt
-
\frac{d}{dx}\int_{0}^{-x} f(t)\;dt
\\
&= 
f(x)- f(-x)
\end{align*}$
But, and here's where I pull my hair out looking over and over for a sign error above or below, from the Leibniz integral rule, we can write:
$f_1(x) = -x \Rightarrow f^\prime_1(x) = -1$
$f_2(x) = x \Rightarrow f^\prime_2(x) = 1$
And so:
$\frac{d}{dx} 
\left(\int_{-x}^{x} g(t) \,dt \right )
= g(x) + g(-x)$
Two conflicting results! Where oh where have I gone wrong? I'm sure I'll be kicking myself soon if you help me spot it - which will be a step up from pulling my hair out.
 A: The fundamental theorem of calculus write
$$\frac d {dx}\int_{a(x)}^{b(x)} f(t) \, dt=f(b(x))\, b'(x)-f(a(x))\, a'(x)$$
A: If you write $F (x)=\int_a^xf (t)\,dt $, you are saying that $$\tag1( F(-x))'=F'(-x). $$ Which is wrong. For instance if $F (x)=x^2$, the equality  $(1) $ becomes $2x=2 (-x) $. 
When you write $F (-x) $, you have $F (h (x))$, with $h (x)=-x $. The derivative, obtained with the chain rule, is 
$$\tag2 (F (h (x))'=F'(h (x))\,h'(x). $$ That's where your missing $-1$ comes from.
A: Let $F(x)$ be an antiderivative of $f(x)$. Then
$$\int_{-x}^xf(t)\,dt=\left.F(t)\right|_{t=-x}^x=F(x)-F(-x).$$
Now by the chain rule,
$$\frac d{dx}\int_{-x}^xf(t)\,dt=(x)'f(x)-(-x)'f(-x)=f(x)+f(-x).$$
A: user539887 found where the error was and forced a major rethink and focus on that one line, from which I finally realised, that I had abused equation 1 when I subsituted -x for x.
The complete substitution yields:
$\frac{d}{-dx}\int_{a}^{-x} f(t) dt = f(-x)$
which can be rearranged to:
$\frac{d}{dx}\int_{a}^{-x} f(t) dt = -f(-x)$
And so my second method should read:
$\begin{align*}
\frac{d}{dx}\int_{-x}^{x} f(t)\;dt
&= 
\frac{d}{dx}\int_{-x}^{0} f(t)\;dt
+
\frac{d}{dx}\int_{0}^{x} f(t)\;dt
\\
&= 
-\frac{d}{dx}\int_{0}^{-x} f(t)\;dt
+
\frac{d}{dx}\int_{0}^{x} f(t)\;dt
\\
&= 
\frac{d}{dx}\int_{0}^{x} f(t)\;dt
+
\frac{d}{dx}\int_{0}^{-x} f(t)\;dt
\\
&= 
f(x) + f(-x)
\end{align*}$
and it agrees with the Leibniz integration rule. All confusion abated. 
