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...
The Chain rule: Derivatives under 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
\( f\circ f^{-1}(x) = f(f^{-1}(x)) = Id = x \)
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.
\( f\circ f^{-1}(x) = f(f^{-1}(x)) = Id = x \)
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
\( f'(x)=(h\circ {g})'(x)= h'(g(x)) g'(x)=h'(x^{2}) 2x \)
\( f'(x)=2x sin^{3}(x^{2}) \)
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
\(f'(x) = (1)-(2) = \frac{6x^{2}}{1+\cos^{3}(x^{3})}-\frac{4x}{1+\cos^{3}(x^{2})} \)
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
\(f'(x) = (1)-(2) = \frac{6x^{2}}{1+\cos^{3}(x^{3})}-\frac{4x}{1+\cos^{3}(x^{2})} \)