differentiate integral in t by x can somebody explain to me why the derivative of
$ h(x) = \int_1^x \ln\lvert \cos(e^\sqrt{t}) + 2 \rvert dt $
is:
$ h(x)' = \ln\lvert \cos(e^\sqrt{x}) + 2 \rvert $
Why does only the variable change?
Thanks
 A: This is the fundemental theorem of calculus. Lets just say that in general we have some function $f(x)$ with antiderivative $F(x)$. This means that
$$
\frac{d}{dx}F(x) = f(x), \: \int f(x)dx = F(x)+C
$$
so if we have that our integral is the following, we can do some steps and derive what you have
$$
\frac{d}{dx}\int_a^x f(t)dt = \frac{d}{dx}\left(F(t)|^x_a\right) = \frac{d}{dx}\left(F(x)-F(a)\right)
$$
now $F(a)$ is just a constant, but look above at what the derivative of $F(x)$ is, its just the original function!
$$
\frac{d}{dx}\left(F(x)-F(a)\right) = f(x)
$$
A: Any integral of the form $$\int_0^x f(t)dt$$ evaluates to $$ F(t) \bigg |_0^x=F(x)-F(0) $$(assuming $F$ is the anti derivative of $f$)
Differentiating this w.r.t $x$ gives $$F'(x)-0 $$ But $F'(x)$ is simply $f(x)$, and we are left with $$f(x)$$
A: You have to perspectives to this problem. Let's make this a bit more general. Take $f$ continuous and $g(x) := \int_a^x f(t)\;dt$.
i) You use your educated guess about $g' = f$. From the fundamental theorem of analysis, you know that
$$
g(x) = \int_a^x f(t)\;dt = F(x) - F(a),
$$
where $F$ is a differentiable function such that $F' = f$. Differentiating $g$ gives
$$
g'(x) = F'(x) - 0 = f(x),
$$
because $F(a)$ is a constant.
ii) You calculate the derivative explicitly. You have
$$
g(x + h) - g(x) = \int_a^{x+h} f(t)\;dt - \int_a^x f(t)\;dt = \int_x^{x+h} f(t)\;dt.
$$
The very famous Lebesgue's density theorem gives you
$$
\dfrac{1}{h} \int_{x-h/2}^{x+h/2} f(t)\;dt \underset{h\downarrow 0}{\to} f(x).
$$
That gives you the result. (Since we are looking at $f$ continuous only, you can also get this result directly without passing through Lebesgue integration. If you haven't seen Lebesgue's density theorem, you might want to try that.)
