TITLE: What's the intuition behind this chain rule usage in the fundamental theorem of calc?
QUESTION [3 upvotes]: I understand this:

but I don't understand this example fully:

So I get the intuition behind the idea that the sliver at some point x in the area function (g(x)) is just the y coordinate of the original function (f(x)), but I'm sure why we need chain rule in the example. Can someone show me an easier example that might pump my intuition?
Like say the example was this instead:
$$ \frac{d}{dx} \int_{a}^{x^2} t $$
So the original function t is just a line with a slope of 1 going up at a 45 degree angle. (1,1) and (2,2) are points on $t$. Using chain rule, the derivative of this graph is going to be:
$$x^2 \cdot 2x = 2x^3$$ right?
And I guess the intuition behind this is that the upper limit (x^2) is increasing exponentially relative to x and so that needs to be taken into account somehow. Chain rule is that way? It's just strange to me that the area isn't just increasing by $$x^2$$ but by $$x^2 \cdot 2x$$

REPLY [1 votes]: In general, let $H'(t)=h(t)$, then:
$$\int_{f(x)}^{g(x)} h(t)dt=H(t)|_{f(x)}^{g(x)}=H(g(x))-H(f(x)) \Rightarrow \\
\begin{align}\left(\int_{f(x)}^{g(x)} h(t)dt\right)'_x&=[H(g(x))-H(f(x))]'_x =\\
&=h(g(x))\cdot (g(x))'_x-h(f(x))\cdot (f(x))'_x.\end{align}$$
Since $f(x)=1,g(x)=x^4, h(t)=\sec t$, then:
$$\left(\int_{1}^{x^4} \sec t \ dt\right)'_x=h(x^4)\cdot (x^4)'_x-h(1)\cdot (1)'_x=\sec x^4\cdot 4x^3.$$
Source: See Variable limits form section at Leibniz integral rule
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