# What's the intuition behind this chain rule usage in the fundamental theorem of calc?

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?

$$\frac{d}{dx} \int_{a}^{x^2} t$$

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 $$x^2 \cdot 2x$$

• Let $F(x)=\int_a^{g(x)} f(t)\,dt$ where $f$ is continuous and $g$ is differentiable. Then, $F(x+h)-F(x)=\int_{g(x)}^{g(x+h)}f(t)\,dt$. Using the mean value theorem, there is a number $\xi\in[g(x),g(x+h)]$ so that $$F(x+h)-F(x)=f(\xi) (g(x+h)-g(x))$$ Now divide by $h$ and let $h\to0$. Can you finish? Does this provide the intuition you covet? – Mark Viola Dec 30 '18 at 20:27
• nah I don't see it. I liked the accepted answer better. – Jwan622 Jan 4 at 14:47
• If you don't understand this, I suggest you have another look and try your best. It is a very basic development. When you divide by $h$ and let $h\to0$ you will get$f(g(x))\,\times\, g'(x)$ – Mark Viola Jan 4 at 17:24

I will use the notation in your screenshot. If you understand that $$g(x) = \int_a^x f(t) \, dt$$ has derivative $$g'(x) = f(x)$$, then $$\int_a^{h(x)} f(t) \, dt = g(h(x))$$ has derivative $$g'(h(x)) h'(x)$$ by the chain rule.
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 .