  # The Chain rule: Derivatives under integral sign

### Basic concepts and principles

A composition of two functions is the operation given by applying a function, then the other one. This operation is denoted by fog(x) ie:

$f\circ g(x) = f(g(x))$

For example, given

$\begin{matrix} f(x)=x^2 \\ g(x)=sin x \end{matrix}$

Then

$f\circ g(x) = f(g(x)) = f(sin(x)) = (sin x)^2$

Chain rule is the formula to define derivative of two functions composition and it is given by

$(f\circ {g})'(x)=lim_{h\rightarrow 0}{\frac{f\circ {g}(x + h) - f\circ {g}(x)}{h}} = f'(g(x)) g'(x)$

This is a basic rule of analysis, it allows to have the inverse derivate formula, derivation under integral sign and many others...

### Inverse function formula

Given a differentiable function f (x), we asked for the derivative of the inverse. We start from

$f\circ f^{-1}(x) = f(f^{-1}(x)) = Id = x$

Then we can apply the chain rule to obtain

$f(f^{-1}(x))' = f'(f^{-1}(x))(f^{-1})'(x) = 1$

Then if f'(x) ≠ 0 , we have

$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$

By the medium value theorem, the fact f'(x) ≠ 0 implies that f is injective, because if there exists x, y distincts such us $f(x) = f(y)$ then we have

$f(x)-f(y) = 0 \Rightarrow \exists z : f'(z)=\frac{f(y)-f(x)}{y-x} =0$ which would be a contradiction.

### Inverse function theorem

Given a differentiable function f such that f'(x0) ≠ 0 then there exists a neighborhood centered at x0 wich we denote by $\mathbb{V}(x_{0})$ , such us
$$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}\, \ \forall x \in \mathbb{V}(x_{0})$$

### Differentiation under the integral sign

In analysis many functions are defined under an integral sign, for example

$f(x)=\int_{a}^{x^2} \frac{1}{1+cos 3t} dt$

We can consider it as composition

$h(x)= \int_{a}^{x}\frac{1}{1+cos3t} dt$

$g(x) = x^{2}$

We can now apply the chain rule to calculate the derivative of f, ie

$f'(x) = (h \circ {g})'(x)= h'(g(x)) g'(x)= h'(x^{2}) 2x=\frac{2x}{1 + cos3x^{2}}$

### Example 1 of Derivatives under integral sign

$f(x)=\int_{a}^{x^{2}}\sin^{3}t dt$

Then as in our example above we take

$h(x)=\int_{a}^{x}\sin^{3}t dt$

$g(x)=x^{2}$

So

$f'(x)=(h\circ {g})'(x)= h'(g(x)) g'(x)=h'(x^{2}) 2x$

$f'(x)=2x sin^{3}(x^{2})$

### Example 2 of Derivatives under integral sign

$f(x)=\int_{x^2}^{x^{3}}\frac{2}{1+\cos^{3}t} dt$

Note that we can split this integral as a sum of two

$\int_{x^2}^{x^{3}}\frac{2}{1+ \cos^{3}t} dt = \int_{-\infty}^{x^{3}}\frac{2}{1+ \cos^{3}t} dt - \int_{-\infty}^{x^{2}}\frac{2}{1+\cos^{3}t} dt = (1) - (2)$

We calculate the first one

$h(x)=(F\circ {G})(x)$ Where

$F(x)=\int_{-\infty}^{x}\frac{2}{1+\cos^{3}t} dt$
$G(x)=x^{3}$

So

$(1) = \frac{6x^{2}}{1+\cos^{3}(x^{3})}$

Now calculate the second one of integrals, again we have

$h(x)=(F\circ {G})(x)$ where

$F(x)=\int_{-\infty}^{x}\frac{2}{1+\cos^{3}t} dt$
$G(x)=x^{2}$

and so

$(2) = \frac{4x}{1+\cos^{3}(x^{2})}$

We already have the result as

$f'(x) = (1)-(2) = \frac{6x^{2}}{1+\cos^{3}(x^{3})}-\frac{4x}{1+\cos^{3}(x^{2})}$

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