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:

fog(x)= f(g(x))

For example, given

f(x)=x2

g(x)= sin x

Then

fog(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

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

# The Chanin rule: Derivatives within the integral sign

### Basic concepts and principles

### Exthended Theory

### Inverse function formula

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

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) = [ f'(f-1(x)) ] -1

By the medium value theorem, fact f'(x) ≠ 0 implies that f is injective, because

f(x)-f(y) = 0 => ∃ z : f'(z) = f(y)-f(x)/(y-x) = 0, which would be a contradiction.

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) = [ f'(f-1(x)) ] -1

By the medium value theorem, fact f'(x) ≠ 0 implies that f is injective, because

f(x)-f(y) = 0 => ∃ z : f'(z) = 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 V(x0), such us

(f-1)'(x) = [ f'(f-1(x)) ] -1 ∀ x ∈ V(x0).

### Differentiation under the integral sign

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

We can consider it as the composition

Where

And

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

### Examples of Derivatives under integral sign

**1)**

Then as in our example above we

So

**2)**

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

We calculate the first one

Where

So

Now calculate the second of integrals, again wew have

Donde

Así

We already have the result as

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