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...

Exthended Theory

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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