$FTC$ problem $\frac d{dx}\int_1^\sqrt xt^tdt$ I stumbled upon a sneaky example while reading about the Fundamental Theorem of Calculus here
$$\frac d{dx}\int_1^\sqrt xt^tdt$$
and it makes me question my whole understanding.
I know that the FTC says 
$$\frac d{dx}\int_a^x f(t)dt\,=\, f(x)$$
so I would say that $f(x)=\sqrt x^\sqrt x$ but this is incorrect I should get
$$f(x)=\frac 12 x^{\frac {\sqrt x}2-\frac 12}$$
which I see as 
$$f(x) = \frac 12 \left(\frac 1{\sqrt x}\right)x^{\frac {\sqrt x}2}$$
I am confused, why does this look like the result I get from $\frac d{dx}\sqrt x^\sqrt x$ and not $\sqrt x^\sqrt x$? I can almost get to $\frac 12 x^{\frac {\sqrt x}2-\frac 12}$ when I take the derivative of $\sqrt x^{\sqrt x}$, I let $u=\sqrt x$ such that
$$\frac d{dx}x^u\cdot\frac {du}{dx} = \left(\frac {\sqrt x}2 x^{\frac{\sqrt x}2-1}\right)\left(\frac 1{2\sqrt x}\right)$$ 
$$\frac d{dx} x^{\frac {\sqrt x}2} = \frac 14 x^{\frac {\sqrt x}2-1}$$
I want to make sure I grasp those concepts before I move on and not just blindly apply formulas so I would like to know what is really going on. Thanks in advance for your help, this is greatly appreciated!
 A: This is the Fundamental Theorem of Calculus, but with a chain rule added. To see this, consider:
$$F(x)=\int_1^xt^tdt$$
You're interested in the derivative of $F(\sqrt x)$, so:
$$\frac{d}{dx}F(\sqrt x) = F'(\sqrt x)*(\sqrt x)'$$
The first term gives you what you have, and the second term will be the term you're missing
A: Let $F$ be a primitive of $f$. Then you have
$$
\int_a^{\sqrt{x}} f(t)dt=F(\sqrt{x})-F(a)
$$
and
$$
F'(x)=f(x).
$$
So you have
$$
\frac{d}{dx}\int_{a}^{\sqrt{x}}t^tdt=\frac{d}{dx}(F(\sqrt{x})-F(a))=\frac{1}{2\sqrt{x}}F'(\sqrt{x})=\frac{1}{2\sqrt{x}}(\sqrt{x})^\sqrt{x}.
$$
Where the second inequality is the derivative of composite function.
A: We can make a substitution here.  Let's write $u=t^2$, so that $du=2tdt$.  Then
\begin{equation}
\int_1^\sqrt{x} t^tdt = \int_1^\sqrt{x}(t^2)^{(t-1)/2}tdt = \frac 12\int_1^x u^{(\sqrt{u}-1)/2}du.
\end{equation}
Applying the FToC to this last integral should give the desired result.  It's important to understand that we've used the chain rule here.  We can run a similar argument for any derivative of the form
\begin{equation}
\frac{d}{dx}\int_a^{g(x)} f(t)dt,
\end{equation}
where $g$ is some one-to-one function.
A: Essentially when you have problems like these where the upper limit is a function, what you do is substitute the variable you are integrating with, in this case $t$, with that function, and then take the derivative.
Thus,
$$\frac d{dx}\int_1^\sqrt xt^tdt = \sqrt{x}^\sqrt{x}*\frac{d}{dx}\sqrt{x}=\frac{\sqrt{x}^\sqrt{x}}{2\sqrt{x}}$$
$$Yuck.$$
A: You could do the transformation
$dx = 2\sqrt{x}d\sqrt{x}$.
And the final result is 
\begin{equation}
\frac{d}{dx}\int_1^{\sqrt{x}}t^tdt = \frac{1}{2\sqrt{x}}\frac{d}{d\sqrt{x}}\int_1^{\sqrt{x}}t^tdt=\frac{\sqrt{x}^{\sqrt{x}}}{2\sqrt{x}}.
\end{equation}
