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.

$$ \forall f \in F(\mathbb{R}, \mathbb{R}), \ \forall \lambda \in \mathbb{R},$$

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

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

$$ \forall (f,g) \in F(\mathbb{R}, \mathbb{R})^2,$$

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

In the same way,

$$ \forall (f,g) \in F(\mathbb{R}, \mathbb{R})^2,$$

$$ \bigl( f \textcolor{#A65757}{-} g \bigr)' = f' \textcolor{#A65757}{-} g' $$

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

$$ \forall (f,g) \in F(\mathbb{R}, \mathbb{R})^2, \ \forall (\lambda, \mu) \in \hspace{0.05em} \mathbb{R}^2 $$

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

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

1. Generalization

$$ \forall n \in \mathbb{N}, \enspace \forall (f,g) \in F(\mathbb{R}, \mathbb{R})^2, \enspace \forall k \in [\![ 1, n ]\!], \enspace \forall \lambda_k \in \hspace{0.05em} \mathbb{R}^n,$$

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

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

1. Simple derivation

$$ \forall (f,g) \in F(\mathbb{R}, \mathbb{R})^2,$$

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



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::

$$ \forall n\in \hspace{0.05em} \mathbb{N}, \enspace \forall (f,g) \in F(\mathbb{R}, \mathbb{R})^2, $$

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

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

$$ \forall g \in F\Bigl( \mathbb{R} , \ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr), $$

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

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

$$ \forall f \in F(\mathbb{R}, \mathbb{R}), \ \forall g \in F\Bigl( \mathbb{R} , \ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr), $$

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

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:

$$ \forall (f,g) \in F(\mathbb{R}, \mathbb{R})^2,$$

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

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,

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

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

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

And under its developped form,

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

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

Recap table of the derivatives of composite functions

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

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

$$ f : I \longmapsto f(I) = J , \enspace x \longmapsto f(x) $$

We define its reciprocal function by:

$$ f^{-1} : J \longmapsto I , \enspace f(x) \longmapsto x $$

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

$$ \forall (f,f^{-1}) \in F(\mathbb{R}, \mathbb{R})^2, \enspace (f' \circ f^{-1}) \neq 0, $$

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

Recap table of the derivatives of operations on functions

Click on the title to access to the recap table.

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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:

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

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

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

So:

$$ \forall f \in F(\mathbb{R}, \mathbb{R}), \ \forall \lambda \in \mathbb{R},$$

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

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

1. Using the limit of the variation rate

With the definition of the derivative, we do have:

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

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

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




$$ \forall (f,g) \in F(\mathbb{R}, \mathbb{R})^2,$$

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




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

$$ \bigl( f \textcolor{#A65757}{-} g \bigr)' = f' \textcolor{#A65757}{-} g' $$




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,

$$ \forall (f,g) \in F(\mathbb{R}, \mathbb{R})^2,$$

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

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

With the derivative of a sum function,

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

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

$$ \forall (f,g) \in F(\mathbb{R}, \mathbb{R})^2, \ \forall (\lambda, \mu) \in \hspace{0.05em} \mathbb{R}^2, $$

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

1. Generalization

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

$$ \forall n \in \mathbb{N}, \enspace \forall (f,g) \in F(\mathbb{R}, \mathbb{R})^2, \enspace \forall k \in [\![ 1, n ]\!], \enspace \forall \lambda_k \in \hspace{0.05em} \mathbb{R}^n,$$

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

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:

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




And finally,

$$ \forall (f,g) \in F(\mathbb{R}, \mathbb{R})^2,$$

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




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,

$$ \forall (f,g) \in F(\mathbb{R}, \mathbb{R})^2,$$

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




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:

$$ \forall (f,g) \in F(\mathbb{R}, \mathbb{R})^2,$$

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

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.

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

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

$$ (fg)^{(k+1)} = \textcolor{#606B9E}{\binom{k}{0}} \left ( f^{(k+1)} \hspace{0.1em} g^{(0)} \right) + \textcolor{#446e4f}{\left[\binom{k}{0} + \binom{k}{1}\right]} \left (f^{(k)} \hspace{0.1em} g^{(1)} \right) + \textcolor{#A65757}{\left[\binom{k}{1} + \binom{k}{2}\right]} \left (f^{(k -1 )} g^{(2)} \right) \enspace + ... + \enspace \textcolor{#7C578A}{\left[\binom{k}{k-1} + \binom{k}{k}\right]} \left (f^{(1)} \hspace{0.1em} g^{(k)} \right) + \textcolor{#606B9E}{\binom{k}{k}} \left ( f^{(0)} \hspace{0.1em} g^{(k+1)} \right) $$

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,

$$ \forall n\in \hspace{0.05em} \mathbb{N}, \enspace \forall (f,g) \in F(\mathbb{R}, \mathbb{R})^2, $$

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

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,

$$ \forall g \in F\Bigl( \mathbb{R} , \ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr), $$

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




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,

$$ \forall g \in F\Bigl( \mathbb{R} , \ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr), $$

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

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, we obtain that:

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

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

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




And finally,

$$ \forall f \in F(\mathbb{R}, \mathbb{R}), \ \forall g \in F\Bigl( \mathbb{R} , \ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr), $$

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

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,

$$ \forall f \in F(\mathbb{R}, \mathbb{R}), \ \forall g \in F\Bigl( \mathbb{R} , \ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr), $$

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

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:

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

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

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

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

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

The limit of a product being the limit product, we can now write:

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

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,

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

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

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




And as a result,

$$ \forall (f,g) \in F(\mathbb{R}, \mathbb{R})^2,$$

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

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,

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

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

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

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 \(f(u(x))\).

$$x \longmapsto u(x) \longmapsto f(u(x))$$

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:

$$ f : I \longmapsto f(I) = J , \enspace x \longmapsto f(x) $$

We define its reciprocal function by:

$$ f^{-1} : J \longmapsto I , \enspace f(x) \longmapsto x $$

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

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

$$ (f \circ f^{-1 })(x) = (f^{-1})'(x) .(f' \circ f^{-1})(x) \qquad (1) $$

So:

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

$$ (f \circ f^{-1 })'(x) = 1 \qquad (2) $$

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

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

And finally,

$$ \forall (f,f^{-1}) \in F(\mathbb{R}, \mathbb{R})^2, \enspace (f' \circ f^{-1}) \neq 0, $$

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

Recap table of the derivatives of operations on functions