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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!

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    $\begingroup$ I don't want to make a whole answer for this, but I think the problem is that you are forgetting a chain rule here. $\frac{d}{dx} \int_a^{g(x)} f(t)dt = \frac{d}{dx} F(g(x)) - F(a) = f(g(x))\cdot g^\prime (x)$. $\endgroup$ – AlexanderJ93 Feb 27 '17 at 23:12
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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

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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.

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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.

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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,

$$\frac d{dx}\int_1^\sqrt xt^tdt = \sqrt{x}^\sqrt{x}*\frac{d}{dx}\sqrt{x}=\frac{\sqrt{x}^\sqrt{x}}{2\sqrt{x}}$$

$$Yuck.$$

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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}

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