How does fundamental theorem of calculus and chain rule work?

Question

I came across a problem of fundamental theorem of calculus while studying Integral calculus.

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A problem:

$\frac{d}{dx}\int_{\pi}^{x^2}\cot^2t\ dt$

which was salved as :

Step I : Let, F(x) = $\int_{\pi}^{x^{ }}\cot^2t\ dt$ ⇒ $F'(x)= \frac{d}{dx}\int_{\pi}^{x^{ }}\cot^2t\ dt=cot^2(x)$

Step II : $\frac{d}{dx}\int_{\pi}^{x^2}\cot^2t\ dt$ = $\frac{d}{dx}[F(x^2)]$ = $F'(x^2)*\frac{d}{dx}(x^2)$

Step III : $F'(x^2)*\frac{d}{dx}(x^2) = cot^2(x^2)*2x$

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I don't understand how can they salve $F'(x^2)$ in third step by putting $x^2$ directly into $F'(x)=cot^2x$

Reference : problem video

Answer 1

I think you're confusing $F'(x^2)$ and $(F(x^2))'$. The first is the function $F'$ $\bf{evaluated}$ at $x^2$ and the second is the derivative of the function $x \mapsto F(x^2).$ These are two different things !

If you take the function $x \mapsto F(x) = 2x + 1$. Then $F'(x) = 2$ so $F'(x^2) = 2$ but $$(F(x^2))' = (x^2)' \cdot F'(x^2) = 2x \cdot 2 = 4x.$$

Similarly for any differentiable function $h$, $h'(2)$ is not necessarily equal to $0$ since $$h'(2) \neq (h(2))' = 0.$$

Answer 2

The problem is totally nonsensical since $$F(u):=\int_\pi^u \cot^2 t\>dt$$ is undefined for any $u\ne\pi$. Apart from this the treatment suffers from using the letter $x$ in "different worlds". In order to make this clear I'm dealing with the function $$G(u):=\int_{\pi/2}^u \cot^2 t\>dt$$ instead (I have replaced $\pi$ by ${\pi\over2}$). This function $u\mapsto G(u)$ is then defined for $\left|u-{\pi\over2}\right|<{\pi\over2}$. One has $$G'(u)=\cot^2 u\ ,\tag{1}$$ by the FTC. You would then be interested in the function $$f(x):=\int_{\pi/2}^{x^2}=G(x^2)$$ and want to know $f'(x)$. By the chain rule and $(1)$ one has $$f'(x)=G'(x^2)\cdot{d\over dx}x^2=\cot^2(x^2)\cdot(2x)\ ,$$ as stated at your source.