The derivatives of operations on functions

Let be by default two functions \( f, g \), depending on \( x \) such as:

$$ \Biggl \{ \begin{align*} \forall x \in D_f, \enspace f: x \longmapsto f(x) \\ \forall x \in D_g, \enspace g: x \longmapsto g(x) \end{align*} $$

Function multiplied by a constant\(: (\lambda f )' \)

When we derive a function mutliplied by a constant \(\lambda \in \mathbb{R}\), we can take it out and derive the function separately.

$$ (\lambda f)' = \lambda f' $$

Sum of two functions\(: (f+g )' \)

$$ \bigl( f + g \bigr)' = f' + g' $$

In the same way,

$$ \bigl( f \textcolor{#8E5B5B}{-} g \bigr)' = f' \textcolor{#8E5B5B}{-} g' $$

Linear combination of two functions\(: (\lambda f+ \mu g )' \)

$$ (\lambda f+ \mu g )' = \lambda f'+ \mu g' $$

The derivative of a linear combination is the linear combination of each derivative function.

1. Generalization
$$ \forall n \in \mathbb{N}, \enspace \forall (f,g), \enspace \forall k \in [\![ 1, n ]\!], \enspace \forall \lambda_k \in \hspace{0.05em} \mathbb{R}^n,$$

$$ \Biggl( \sum_{k=0}^n \lambda_k f_k \Biggl)' = \sum_{k=0}^n \lambda_k f'_k $$

Product of two functions\(: (fg )' \)

1. Simple derivation
$$ \forall (f,g),$$

$$ \left ( f g \right)' = f'g + g'f $$



2. Multiple derivation: the Leibniz's formula

Let be \(f, g\) two functions of class \( \mathbb{C}^{\infty}\) on an interval \(I\). We note \(f^{(n)}\) the \(n\)-th derivative of \(f\).




The Leibniz's formula tells us that::

$$ \forall n\in \hspace{0.05em} \mathbb{N}, \enspace \forall (f,g), $$

$$ (fg)^{(n)} = \sum_{p = 0}^n \binom{n}{p} f^{(n-p)} \hspace{0.1em} g^{(p)} \qquad (Leibniz ) $$

Inverse of a function\(: (1 / g )' \)

$$ \left ( 1 \over g \right)' = \frac{g'}{g^2} $$

Quotient of two functions\(: (f / g )' \)

$$ \left ( f \over g \right)' = \frac{f'g - g'f}{g^2} $$

Composite functions\(: (f \circ g )' \)

1. Two-functions composite

Let be \( f, g \) two functions.

$$ g : I \longmapsto J , \enspace x \longmapsto g(x) $$
$$ f : J \longmapsto K, \enspace y = g(x) \longmapsto f(y) = f \left(g(x)\right) $$

We define a composite function \( (f \circ g) \) as:

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



Its derivative is:

$$ \forall (f,g),$$

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

So:

$$ (f \circ g)' = g'.f' \left(g\right) $$

We also call it a chain derivation.



2. Generalization: any composite function

Let us define a new operator of composition:

$$ \forall n \in \mathbb{N}, \ \Biggl( \overset{n}{\underset{k=1}{\bigcirc f_k}} \ \Biggr )(x) = \Bigl(f_1 \circ f_2 \circ f_3 \circ \ ... \ \circ f_{n-1} \circ f_{n}\Bigr)(x) $$

We now define a new function \(\Psi_n (x) \):

$$ \forall n \in \mathbb{N}, \ \Psi_n (x) = \ \Biggl( \overset{n}{\underset{k=1}{\bigcirc f_k}} \ \Biggr ) (x) $$



Thus we can model it as,

$$ \forall n \in \mathbb{N}, \ \forall k \in [\![1, n ]\!], \ \forall f_k \in \hspace{0.05em} f^k,$$

$$ \Psi_n' = f'_n \times \prod_{k=1}^{n-1}\Biggl[ f'_{n-k} \circ \Biggl( \overset{n}{\underset{j= (n - k) + 1}{\bigcirc f_j}} \ \Biggr ) \Biggr] \\ $$

$$ with \enspace \Psi_n (x) = \Biggl( \overset{n}{\underset{k=1}{\bigcirc f_k}} \ \Biggr ) (x) = \Bigl(f_1 \circ f_2 \circ f_3 \circ \ ... \ \circ f_{n-1} \circ f_{n}\Bigr)(x) $$

And under its developped form,

$$ \forall n \in \mathbb{N}, \ \forall k \in [\![1, n ]\!], \ \forall f_k \in \hspace{0.05em} f^k,$$

$$ \Psi_n' = \Biggl( \overset{n}{\underset{k=1}{\bigcirc f_k}} \ \Biggr )'= f_n' \times \bigl( f_{n-1}' \circ f_n \bigr) \times \ ... \ \times \bigl(f_1' \circ f_2 \circ \ ... \ \circ f_{n-1} \circ f_n \bigr)$$

Recap table of the derivatives of composite functions

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Reciprocal function\(: (f^{-1} )' \)

Let be \( f \) a function such as:

We define its reciprocal function by:

The reciprocal function derivative \( (f^{-1})' \) is:

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

Recap table of the derivatives of operations on functions

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Demonstrations

Function multiplied by a constant\(: (\lambda f )' \)

Let be \( \lambda \in \mathbb{R} \) a real number.

With the definition of the derivative, we do have:

We can factorize it by \( \lambda \):

So:

$$ (\lambda f)' = \lambda f'$$

Sum of two functions\(: (f+g )' \)

1. Using the limit of the variation rate

With the definition of the derivative, we do have:

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

The limit of a sum being the sum of both limits:

$$ lim (f+g) = lim (f) + lim (g) $$
$$ ( f+ g)'(x) = lim_{h \to 0 } \enspace \frac{ f(x+h) - f(x)}{h} + lim_{h \to 0 } \enspace \frac{g(x+h) - g(x)}{h} $$


$$ \forall (f,g),$$

$$ \bigl( f + g \bigr)' = f' + g'$$




In the same way, a difference being a sum of negative number, it follows that:

$$ \bigl( f \textcolor{#8E5B5B}{-} g \bigr)' = f' \textcolor{#8E5B5B}{-} g' $$




2. Using partial derivatives

Considering \( y \) as a function depending on two variables \(f \) and \( g \):

$$y = f + g$$

We thus have a partial derivative:

$$ dy = \frac{\partial y}{\partial f} df + \frac{\partial y}{\partial g}dg $$
$$ dy = df + dg $$

And then derivating now in relation to \(x \):

$$ \frac{dy}{dx} = \frac{df}{dx} + \frac{df}{dx} $$



As a result we do have,

$$ \forall (f,g),$$

$$ \bigl( f + g \bigr)' = f' + g'$$

Linear combinations of two functions\(: (\lambda f+ \mu g )' \)

With the derivative of a sum function,

Finally, with the derivative of a function multiplied by a constant \( \lambda \), we can directly conclude that:

$$ (\lambda f+ \mu g )' = \lambda f'+ \mu g' $$

1. Generalization

As well, repeating this operation multiple times, we can establish that:

$$ \forall n \in \mathbb{N}, \enspace \forall (f,g), \enspace \forall k \in [\![ 1, n ]\!], \enspace \forall \lambda_k \in \hspace{0.05em} \mathbb{R}^n,$$

$$ \Biggl( \sum_{k=0}^n \lambda_k f_k \Biggl)' = \sum_{k=0}^n \lambda_k f'_k $$

Product of two functions\(: (fg )' \)

1. Simple derivation

1. Using the limit of the variation rate

With the definition of the derivative, we do have:

$$ \left ( fg \right)'(x) = lim_{h \to 0 } \enspace \frac{ f(x+h)g(x+h) - fg}{h} $$

Let us add the term \( f(x + h)g(x) \), then let us remove it straight away, to preserve the integrity of our initial equation:

$$ \left ( fg \right)'(x) = lim_{h \to 0 } \enspace \frac{ f(x+h)g(x+h) + f(x + h)g(x) - f(x + h)g(x) - fg}{h} $$

We now factorize it by \( f(x + h) \) and by \( g(x) \):

$$ \left ( fg \right)'(x) = lim_{h \to 0 } \enspace \frac{ g(x)(f(x + h) - f(x))}{h} \ + \ lim_{h \to 0 } \enspace \frac{ f(x+h)(g(x+h) - g(x))}{h} $$

The limit of a product being the limit of the product, we now have:

The limit of a product being the limit of the product:

$$ lim (fg) = lim (f) \ lim(g)$$

We now have this:

$$ \left ( fg \right)'(x) = lim_{h \to 0 } \enspace g(x) . \left( lim_{h \to 0 } \enspace \frac{ f(x+h) - f(x)}{h} \right) \ + \ lim_{h \to 0 } \enspace f(x+h) \left( lim_{h \to 0 } \enspace \frac{ g(x+h) - g(x)}{h} \right) $$




And finally,

$$ \forall (f,g),$$

$$ \left ( f g\right)' = f'g + g'f $$




2. Using partial derivatives

Considering \( y \) as a function depending on two variables \(f \) and \( g \):

$$y = fg$$

We thus have a partial derivative:

$$ dy = \frac{\partial y}{\partial f} df + \frac{\partial y}{\partial g}dg $$
$$ dy = g \ df + f \ dg $$

And then derivating now in relation to \(x \):

$$ \frac{dy}{dx} = g \ \frac{df}{dx} + f \frac{dg}{dx} $$

As a result,

$$ \forall (f,g),$$

$$ \left ( f g\right)' = f'g + g'f $$




2. Multiple derivation: the Leibniz's formula

Let be \(f, g\) two functions of class \( \mathbb{C}^{\infty}\) on an interval \(I\). We note \(f^{(n)}\) the \(n\)-th derivative of \(f\).

We know that the derivative of the product of functions is worth:

$$ \forall (f,g),$$

$$ \left ( f g \right)' \hspace{0.1em }= f'g + g'f $$

Moreover, if we derivate it again,

$$ \left ( f g \right)'' \hspace{0.1em } = (f'g)'+ (g'f)' $$
$$ \left ( f g \right)'' \hspace{0.1em } = f''g + g'f' + f'g' + g''f $$
$$ \left ( f g \right)'' \hspace{0.1em } = f''g + 2g'f' + g''f $$

A pattern similar to the Newton's binomial appears in this equation.

Indeed, this can make us think of:

$$ \Biggl \{ \begin{align*} \left ( f g \right)^{(2)} \hspace{0.1em } = f^{(2)}g + 2g'f' + g^{(2)}f \qquad \\ (a+b)^2 = a^2 + 2ab + b^2 \qquad (Newton's \ binomial) \end {align*} $$

Let us derivate it another time,

$$ \left ( f g \right)^{(3)} \hspace{0.1em }= f^{(3)}g + g'f^{(2)} + 2 (g'f^{(2)} + f'g^{(2)}) + g^{(3)}f + f'g^{(2)} $$
$$ \left ( f g \right)^{(3)} \hspace{0.1em }= f^{(3)}g + 3f^{(2)}g' + 3f'g^{(2)} + g^{(3)}f $$

It seems that the successive derivations of the product give:

$$ (fg)^{(n)} = \sum_{p = 0}^n \binom{n}{p} f^{n-p} \hspace{0.1em} g^p $$

Let us try to demonstrate this by recurrence.




1. Proof by recurrence

Let us try to show that the following statement \((S_n)\) is true:

$$ \forall n \in \hspace{0.05em} \mathbb{N}, \enspace (fg)^{(n)} = \sum_{p = 0}^n \binom{n}{p} f^{(n-p)} \hspace{0.1em} g^{(p)} \qquad (S_n) $$

That is to say for all \(k\):

$$ (fg)^{(k)} = \binom{k}{0} f^{(k)} \hspace{0.1em} g^{(0)} + \binom{k}{1} f^{(k-1)} \hspace{0.1em} g^{(1)} + \binom{k}{2} f^{(k-2)} \hspace{0.1em} g^{(2)} \enspace + ... + \enspace \binom{k}{k-1} f^{(1)} \hspace{0.1em} g^{(k-1)} + \binom{k}{k} f^{(0)} \hspace{0.1em} g^{(k)} \qquad (S_{k}) $$

1. Calculation of the first term

Let verify if this is the case for the first term, that is to say when \( n = 0 \).

$$ \sum_{p = 0}^n \binom{n}{p} f^{(n-p)} \hspace{0.1em} g^{(p)} = \binom{n}{0} f^{(0)}g^{(0)} = fg $$

It is just the product \((fg)\) before the derivation.

Thus, \((S_0)\) is true.




2. Heredity

Let \( k \in \mathbb{N} \) be natural number.

Let us assume that \((S_k)\) is true for all \( k \).

$$ (fg)^{(k)} = \sum_{p = 0}^{k} \binom{k}{p} f^{(k-p)} \hspace{0.1em} g^{(p)} \qquad (S_{k}) $$

And let verify if it is also the case for \((S_{k + 1})\).

$$ (fg)^{(k+1)} = \sum_{p = 0}^{k+1} \binom{k+1}{p} f^{(k+1-p)} \hspace{0.1em} g^{(p)} \qquad (S_{k+1}) $$

So that:

$$ (fg)^{(k+1)} = \binom{k+1}{0} f^{(k+1)} \hspace{0.1em} g^{(0)} + \binom{k+1}{1} f^{(k-1)} \hspace{0.1em} g^{(1)} + \binom{k+1}{2} f^{(k-1)} \hspace{0.1em} g^{(2)} \enspace + ... + \enspace \binom{k+1}{k} f^{(1)} \hspace{0.1em} g^{(k)} + \binom{k+1}{k+1} f^{(0)} \hspace{0.1em} g^{(k+1)} \qquad (S_{k+1}) $$




Let start from the expression \((fg)^{(k)}\) and calculate its derivative.

$$ (fg)^{(k)} = \binom{k}{0} f^{(k)} \hspace{0.1em} g^{(0)} + \binom{k}{1} f^{(k-1)} \hspace{0.1em} g^{(1)} \enspace + ... + \enspace \binom{k}{k-1} f^{(1)} \hspace{0.1em} g^{(k-1)} + \binom{k}{k} f^{(0)} \hspace{0.1em} g^{(k)} \qquad (S_{k}) $$

$$ \left ( (fg)^{(k)} \right)' = \binom{k}{0} \left ( f^{(k+1)} \hspace{0.1em} g^{(0)} + f^{(k)} \hspace{0.1em} g^{(1)} \right) + \binom{k}{1} \left (f^{(k)} \hspace{0.1em} g^{(1)} + g^{(2)} f^{(k -1 )} \right) \enspace + ... + \enspace \binom{k}{k-1} \left( f^{(2)} \hspace{0.1em} g^{(k-1)} + g^{(k)}f^{(1)} \right) + \binom{k}{k}\left( f^{(1)} g^{(k)} + g^{(k+1)}f^{(0)} \right) $$

$$ (fg)^{(k+1)} = \textcolor{#606B9E}{\binom{k}{0}} \left ( f^{(k+1)} \hspace{0.1em} g^{(0)} \right) + \textcolor{#446e4f}{\left[\binom{k}{0} + \binom{k}{1}\right]} \left (f^{(k)} \hspace{0.1em} g^{(1)} \right) + \textcolor{#8E5B5B}{\left[\binom{k}{1} + \binom{k}{2}\right]} \left (f^{(k -1 )} g^{(2)} \right) \enspace + ... + \enspace \textcolor{#7C578A}{\left[\binom{k}{k-1} + \binom{k}{k}\right]} \left (f^{(1)} \hspace{0.1em} g^{(k)} \right) + \textcolor{#606B9E}{\binom{k}{k}} \left ( f^{(0)} \hspace{0.1em} g^{(k+1)} \right) $$

But we know thanks to the Pascal's formula, that:

$$ \binom{n}{p} = \binom{n - 1}{p - 1} + \binom{n - 1}{p} \qquad (Pascal) $$

And therefore:

$$ \binom{n + 1}{p + 1} = \binom{n}{p} + \binom{n}{p + 1} \qquad (Pascal^*) $$

So, thanks to \( (Pascal^*) \), we do have now:

$$ (fg)^{(k+1)} = \textcolor{#606B9E}{\binom{k}{0}} \left ( f^{(k+1)} \hspace{0.1em} g^{(0)} \right) + \textcolor{#446e4f}{\binom{k+1}{1}} \left (f^{(k)} \hspace{0.1em} g^{(1)} \right) + \enspace + ... + \enspace \textcolor{#7C578A}{\binom{k+1}{k}} \left (f^{(1)} \hspace{0.1em} g^{(k)} \right) + \textcolor{#606B9E}{\binom{k}{k}} \left ( f^{(0)} \hspace{0.1em} g^{(k+1)} \right) $$

However, we notice that:

$$ \binom{k}{0} = \binom{k + 1}{0} $$

As well as,

$$ \binom{k}{k} = \binom{k + 1}{k + 1} $$

And at the end that:

$$ (fg)^{(k+1)} = \binom{k+1}{0} \left ( f^{(k+1)} \hspace{0.1em} g^{(0)} \right) + \binom{k+1}{1} \left (f^{(k)} \hspace{0.1em} g^{(1)} \right) \enspace + ... + \enspace \binom{k+1}{k} \left (f^{(1)} \hspace{0.1em} g^{(k)} \right) + \binom{k+1}{k+1} \left ( f^{(0)} \hspace{0.1em} g^{(k+1)} \right) $$

We can rewrite this term in the form of sum, and we find our statement \(( S_{k + 1} ) \):

$$ (fg)^{(k+1)} = \sum_{p = 0}^{k+1} \binom{k+1}{p} f^{(k+1-p)} \hspace{0.1em} g^{(p)} \qquad (S_{k+1}) $$

Thus, \((S_{k + 1})\) is true.




3. Conclusion

The statement \((S_n)\) is true for its first terme \(n_0 = 0\) and it is hereditary from terms to terms for all \(k \in \mathbb{N}\).

By the recurrence principle, that statement is true for all \(n \in \mathbb{N}\).




And finally,

$$ \forall n\in \hspace{0.05em} \mathbb{N}, \enspace \forall (f,g), $$

$$ (fg)^{(n)} = \sum_{p = 0}^n \binom{n}{p} f^{(n-p)} \hspace{0.1em} g^{(p)} \qquad (Leibniz ) $$

Inverse of a function\(: (1 /g )' \)

Let \(g \neq 0\) be a non-zero function.

1. Using the limit of the variation rate

With the definition of the derivative, we do have:

$$ \left ( 1 \over g \right)'(x) = lim_{h \to 0 } \enspace \frac{ \frac{1}{g(x+h)} - \frac{1}{g(x)} }{h} $$

Let put the numerator in a common denominator:

$$ \left ( 1 \over g \right)'(x) = lim_{h \to 0 } \enspace \frac{1}{h} . \frac{ g(x) - g(x +h) }{g(x +h)g(x)} $$
$$ \left ( 1 \over g \right)'(x) = lim_{h \to 0 } \enspace \frac{g(x) - g(x +h)}{h} . \frac{ 1 }{g(x +h)g(x)} $$

We now recognize the definition of the derivative of \(g \):

$$ \left ( 1 \over g \right)'(x) = lim_{h \to 0 } \enspace - \frac{g(x +h)- g(x) }{h} . \frac{ 1 }{g(x +h)g(x)} $$
$$ \left ( 1 \over g \right)'(x) = - g'(x) . \frac{ 1 }{g(x)^2} $$



And finally,

$$ \forall g \neq 0, $$

$$ \left ( 1 \over g \right)' = -\frac{ g' }{g^2}$$




2. Using the Leigniz notation

Considering the function \( y\) cas a function depending on the variable \(g \):

$$y = \frac{1}{g}$$
$$ dy = -\frac{1}{g^2} dg $$

So now derivating in relation to \(x \):

$$ \frac{dy}{dx} = -\frac{1}{g^2} \Biggl[ \frac{dg}{dx} \Biggr]$$



As a result,

$$ \forall g \neq 0, $$

$$ \left ( 1 \over g \right)' = -\frac{ g' }{g^2}$$

Quotient of two functions\(: (f / g )' \)

Let \(f\) be a function and \(g \neq 0\) a non-zero function.

1. Using the limit of the variation rate

With the definition of the derivative, we do have:

$$ \left ( f \over g \right)'(x) = lim_{h \to 0 } \enspace \frac{ \frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)} }{h} $$

Let put the numerator in a common denominator:

$$ \left ( f \over g \right)'(x) = lim_{h \to 0 } \enspace \frac{ f(x+h)g(x) - f(x)g(x+h) }{h.g(x).g(x+h)} $$



Let us add the term \( f(x)g(x) \), then let us remove it afterwards, in order to preserve our equation's integrity:

$$ \left ( f \over g \right)'(x) = lim_{h \to 0 } \enspace \frac{ f(x+h)g(x) - f(x)g(x+h) + f(x)g(x) - f(x)g(x)}{h.g(x).g(x+h)} $$

We no factorize it by \( f(x) \) and by \( g(x) \):

$$ \left ( f \over g \right)'(x) = lim_{h \to 0 } \enspace \frac{ g(x).\Bigl[f(x+h) - f(x)\Bigr] - f(x)\Bigl[g(x + h) - g(x)\Bigr]}{h.g(x).g(x+h)} $$



At this stage, let us seperate the expression in two parts:

$$ \left ( f \over g \right)'(x) = lim_{h \to 0 } \enspace \frac{1}{g(x).g(x+h)} \enspace . \biggl[ lim_{h \to 0 } \enspace \biggl( \frac{ g(x).\left(f(x+h) - f(x)\right)}{h} - \frac{ f(x)\left(g(x + h) - g(x)\right)}{h} \biggr) \Biggr] $$

The limit of a difference being the difference of the limit :

$$ lim (f-g) = lim (f)-lim (g) $$

In the expression under brackets we obtain that:

$$ \left ( f \over g \right)'(x) = lim_{h \to 0 } \enspace \frac{1}{g(x).g(x+h)} \enspace . \biggl[ lim_{h \to 0 } \enspace \frac{ g(x).\left(f(x+h) - f(x)\right)}{h} - lim_{h \to 0 } \enspace \frac{f(x)\left(g(x + h) - g(x)\right)}{h} \Biggr] $$

As well, the limit of a product is the product of both limits:

$$ lim (fg) = lim (f) \ lim (g) $$

$$ \left ( f \over g \right)'(x) = lim_{h \to 0 } \enspace \frac{1}{g(x).g(x+h)} \enspace . \biggl[ \Bigl( lim_{h \to 0 } \enspace g(x) \Bigr). lim_{h \to 0 } \enspace \Biggl( \frac{ f(x+h) - f(x)}{h}\Biggr) - \Bigl( lim_{h \to 0 } \enspace f(x) \Bigr).lim_{h \to 0 } \enspace \Biggl( \frac{g(x + h) - g(x)}{h} \Biggr) \Biggr] $$

And in the end, applying this limit, we get this:

$$ \left ( f \over g \right)'(x) = \frac{1}{g(x)^2} \enspace . \biggl( g(x).f'(x) - f(x).g'(x) \biggr) $$



And finally,

$$ \forall (f,g), \ g \neq 0, $$

$$ \left ( f \over g \right)' = \frac{f'g - g'f}{g^2} $$

2. Using partial derivatives

Considering \( y \) as a function depending on two variables \(f \) and \( g \):

$$y = \frac{f}{g}$$

We thus have a partial derivative:

$$ dy = \frac{\partial y}{\partial f} df + \frac{\partial y}{\partial g}dg $$
$$ dy = \frac{1}{g} df - \frac{f}{g^2}dg $$

Putting both terms on the same denominator, we do have:

$$ dy = \frac{g}{g^2} df - \frac{f}{g^2}dg $$
$$ dy = \frac{g.df - f.dg}{g^2} $$

So now derivating in relation to \(x \):

$$ \frac{dy}{dx} = \frac{1}{g^2} \left( g \frac{df}{dx} - f \frac{dg}{dx} \right ) $$



And finally,

$$ \forall (f,g), \ g \neq 0, $$

$$ \left ( f \over g \right)' = \frac{f'g - g'f}{g^2} $$

Composite functions\(: (f \circ g )' \)

1. Two-functions composite

Let be \( f, g \) two functions.

$$ g : I \longmapsto J , \enspace x \longmapsto g(x) $$
$$ f : J \longmapsto K, \enspace y = g(x) \longmapsto f(y) = f \left(g(x)\right) $$

We define a composite function \( (f \circ g) \) as:

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



With the definition of the derivative, we do have:

$$ (f \circ g)'(x) = lim_{h \to 0 } \enspace \frac{(f \circ g)(x + h) - (f \circ g)(x)}{h} $$
$$ (f \circ g)'(x) = lim_{h \to 0 } \enspace \frac{f\bigl(g(x + h) \bigr)- f\bigl(g(x)\bigr)}{h} $$

Let us multiply by a quotient which is worth \(1\), keeping our equation true:

$$ (f \circ g)'(x) = lim_{h \to 0 } \enspace \frac{f\bigl(g(x + h) \bigr)- f\bigl(g(x)\bigr)}{h} . \frac{g(x+ h) - g(x)}{g(x+ h) - g(x)} $$
$$ (f \circ g)'(x) = lim_{h \to 0 } \enspace \frac{f\bigl(g(x + h) \bigr)- f\bigl(g(x)\bigr)}{g(x+ h) - g(x)} . \frac{g(x+ h) - g(x)}{h} $$

The limit of a product being the product of both limits:

$$ lim (fg) = lim (f) \ lim (g) $$

We can now write:

$$ (f \circ g)'(x) = lim_{h \to 0 } \enspace \frac{f(g(x + h))- f\bigl(g(x)\bigr)}{g(x+ h) - g(x)} . lim_{h \to 0 } \enspace \frac{g(x+ h) - g(x)}{h} $$

Let set down a variable change such as:

$$ g(x + h) - g(x) = H $$

When \( h \to 0 \), then \( H \to 0 \).

Moreover, we consider as an hypothesis that:

$$ g(x) = y $$

So,

$$ (f \circ g)'(x) = lim_{H \to 0 } \enspace \frac{f(y + H) - f(y)}{H} . lim_{h \to 0 } \enspace \frac{g(x+ h) - g(x)}{h} $$
$$ (f \circ g)'(x) = f' \left(y\right) . g'(x) $$
$$ (f \circ g)'(x) = f' \bigl(g(x)\bigr) . g'(x) $$



And as a result,

$$ \forall (f,g),$$

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

We also call it a chain derivation.




2. Generalization: any composite function

Let us define a new operator of composition:

$$ \forall n \in \mathbb{N}, \ \Biggl( \overset{n}{\underset{k=1}{\bigcirc f_k}} \ \Biggr )(x) = \Bigl(f_1 \circ f_2 \circ f_3 \circ \ ... \ \circ f_{n-1} \circ f_{n}\Bigr)(x) $$

We now define a new function \(\Psi_n (x) \):

$$ \forall n \in \mathbb{N}, \ \Psi_n (x) = \ \Biggl( \overset{n}{\underset{k=1}{\bigcirc f_k}} \ \Biggr ) (x) $$

So that:

$$ \forall n \in \mathbb{N}, \ \forall k \in [\![1, n ]\!], \ \Psi_k = \left \{ \begin{align*} \Psi_1 = f_1 \\ \Psi_2 = (f_1 \circ f_2 ) \\ \Psi_3 = (f_1 \circ f_2 \circ f_3 ) \\ \Psi_4 = (f_1 \circ f_2 \circ f_3 \circ f_4 ) \\ ... \\ \Psi_n = (f_1 \circ f_2 \circ \ ... \ \circ f_{n-1}\circ f_n ) \end{align*} \right \} $$

Using a chain derivation multiple times, we found out that:

$$ \forall n \in \mathbb{N}, \ \forall k \in [\![1, n ]\!], \ \bigl(\Psi_k \bigl)' = \left \{ \begin{align*} \Psi_1 ' = f_1 ' \\ \Psi_2 ' = f_2' \times (f_1' \circ f_2 ) \\ \Psi_3 ' = f_3' \times \bigl( f_2' \circ f_3 \bigr) \times \bigl(f_1' \circ f_2 \circ f_3 \bigr) \\ \Psi_4' = f_4' \times \bigl( f_3' \circ f_4 \bigr) \times \bigl(f_2' \circ f_3 \circ f_4 \bigr) \times \bigl(f_1' \circ f_2 \circ f_3 \circ f_4 \bigr) \\ ... \\ \Psi_n' = f_n' \times \bigl( f_{n-1}' \circ f_n \bigr) \times \bigl( f_{n-2}' \circ f_{n-1} \circ f_n \bigr) \times \ ... \ \times \bigl(f_1' \circ f_2 \circ \ ... \ \circ f_{n-1} \circ f_n \bigr) \\ \end{align*} \right \} $$




Thus we can model it by,

$$ \forall n \in \mathbb{N}, \ \forall k \in [\![1, n ]\!], \ \forall f_k \in \hspace{0.05em} f^k,$$

$$ \Psi_n' = f'_n \times \prod_{k=1}^{n-1}\Biggl[ f'_{n-k} \circ \Biggl( \overset{n}{\underset{j= (n - k) + 1}{\bigcirc f_j}} \ \Biggr ) \Biggr] \\ $$

$$ with \enspace \Psi_n (x) = \Biggl( \overset{n}{\underset{k=1}{\bigcirc f_k}} \ \Biggr ) (x) = \Bigl(f_1 \circ f_2 \circ f_3 \circ \ ... \ \circ f_{n-1} \circ f_{n}\Bigr)(x) $$

And under a developped form,

$$ \Psi_n' = \Biggl( \overset{n}{\underset{k=1}{\bigcirc f_k}} \ \Biggr )'= f_n' \times \bigl( f_{n-1}' \circ f_n \bigr) \times \ ... \ \times \bigl(f_1' \circ f_2 \circ \ ... \ \circ f_{n-1} \circ f_n \bigr)$$




3. Examples



1. Calculation of \( cos(2x)' \)

To calculate the derivative of \( cos(2x) \), we set down that:

$$ \Biggl \{ \begin{align*} g(x) = 2x \\ f(x) = cos(x) \end{align*} $$

Then, we calculate their respective derivative:

$$ \Biggl \{ \begin{align*} g'(x) = 2 \\ f'(x) = -sin(x) \end{align*} $$

We now apply the following formula:

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

And,

$$ cos(2x)'= -2.sin(2x) $$
2. Calculation of \( \Bigl( \sqrt{e^{x^2}} \Bigr)' \)

Here, we have to considerate a triple chain derivation:

$$ \left \{ \begin{align*} h(x) = x^2 \\ g(x) = e^x \\ f(x) = \sqrt{x} \end{align*} \right \} $$

We calculate their respective derivative:

$$ \left \{ \begin{align*} h'(x) = 2x\\ g'(x) = e^x \\ f'(x) = \frac{1}{2\sqrt{x}} \end{align*} \right \} $$

Applying the simple formula twice in a row or the multiple composite functions give the same result:

$$ (f \circ g \circ h)' = h' (g' \circ h) (f' \circ (g \circ h) ) $$

So:

$$ \Bigl( \sqrt{e^{x^2}} \Bigr)' = 2x e^{x^2} \frac{1}{2\sqrt{e^{x^2}}} $$
$$ \Bigl( \sqrt{e^{x^2}} \Bigr)' = \frac{ x e^{x^2}}{\sqrt{e^{x^2}}} $$
$$ \Bigl( \sqrt{e^{x^2}} \Bigr)' = x \sqrt{e^{x^2}} $$

Recap table of the derivatives of composite functions

This a recap table of composite function of type \(f(u(x))\).

In all these different cases, it will be necessary depending on the intermediate function \(u\), to restrict the domain of definition of \(f(u)\) at most that of \(u\).

Reciprocal function\(: (f^{-1} )' \)

Let be \( f \) a function such as:

We define its reciprocal function by:

Let us start from the previous result of the derivative of a composite function:

Let us compose \( f \) with its reciprocal \( f^{-1} \):

So:

Now injecting the right member of \( (1) \) in the left one of \( (2) \), we obtain:

And finally,

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

Recap table of the derivatives of operations on functions