Differentiation

Derivative

Definition

The derivative quantifies the sensitivity to change of a function's output with respect to its input.

Notations

• For a given function we denote the derivative of a function $f(x)$ to be \( f'(x) \), which is referred to as prime notation.

• For a given function $f$, the derivative $f'$ is often denoted by:

  $$ \frac{d}{dx} f(x) \quad \text{or} \quad \frac{d f(x)}{dx} $$

  This notation is called Leibniz notation.

Differentiability

Definition

We say that \( f : D \to \mathbb{R} \) is differentiable at \( x \) if the limit

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

exists. We call \( f'(x) \) the derivative of \( f(x) \). If \( f'(x) \) exists for all \( x \in D \) then we say \( f \) is differentiable.

Differentiation and Continuity

Theorem

If \( f : D \rightarrow \mathbb{R} \) is differentiable at \( a \in D \) then \( f \) is continuous at \( a \).

Proof

We want to show that \(\lim_{x \to a}f(x) = f(a)\). But if \( x = a + h \) then this is the same as \(\lim_{h \to 0}f(a+h) = f(a)\). We will show that this is true by proving \(\lim_{h \to 0}(f(a+h) - f(a)) = 0\).

\[\lim_{h \to 0}(f(a+h) - f(a)) = \lim_{h \to 0}\frac{f(a+h) - f(a)}{h} \cdot h\] \[= \lim_{h \to 0}\frac{f(a+h) - f(a)}{h} \cdot \lim_{h \to 0}h\] \[= f'(a) \cdot 0\] \[= 0\]

$\square$

Remarks

• Note that this proof only makes sense when \( f'(a) \) exists, that is, when \( f \) is differentiable.

• The converse to this theorem is not true.

Absolute value function

Theorem

The function \( f(x) = |x| \) is continuous, but not differentiable.

Proof

At the point \(a=0\) we have

\[ \lim_{x\to 0^{-}}|x|=0=\lim_{x\to 0^{+}}|x|\quad\text{ so }\quad\lim_{x\to 0}f(x)=0=f(0) \]

So the function is continuous.
We will now show that \(f(x)\) is not differentiable. Unsurprisingly, the interesting thing happens when \(x=0\).

\[ \lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0}\frac{|h|}{h}=\begin{cases} -1 & \text{if }h < 0\\ 1 & \text{if }h>0 \end{cases} \]

The expression \(\left.\frac{|h|}{h}\right|\) is not defined when \(h=0\) (but we don't care about this as \(h\) never equals zero in our limit). So

\[ \lim_{h\to 0^{-}}\frac{|h|}{h}=-1\qquad\text{ and }\qquad\lim_{h\to 0^{+}}\frac{|h|}{h}=1 \]

These limits are not equal, so \(\lim_{h\to 0}\frac{|h|}{h}\) does not exist. So \(f(x)=|x|\) is not differentiable when \(x=0\).

$\square$

Higher derivatives

Definition

Deriving a function $n$ times is called taking the $n$-th derivative.

Notation

As for the prime notation we denote higher derivatives:

$$ f'''^{...}(x) \quad \text{or} \quad f^{(n)}(x) $$

As for the Leibnizian notation we write:

$$ \frac{d^n f(x)}{dx^n} $$

Remark

For some functions it may not be possible or useless to take the derivative more than a few times.

Differentiation of common functions

See the proof of many common functions under démonstration des fonctions dérivées.

Sum rule of differentiation

Theorem

If \(f\) and \(g\) are differentiable at \(x\) then

\[ (f+g)^{\prime}(x) = f^{\prime}(x) + g^{\prime}(x) \]

Proof

We want to show that \((f+g)^{\prime}(x)\) exists and is equal to \(f^{\prime}(x)+g^{\prime}(x)\). Recall that

\[ f^{\prime}(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \]

so

\[ (f+g)^{\prime}(x) = \lim_{h\to 0}\frac{(f+g)(x+h)-(f+g)(x)}{h} \] \[ = \lim_{h\to 0}\frac{f(x+h)+g(x+h)-f(x)-g(x)}{h} \] \[ = \lim_{h\to 0}\frac{(f(x+h)-f(x))+(g(x+h)-g(x))}{h} \] \[ = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} + \frac{g(x+h)-g(x)}{h} \] \[ = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} + \lim_{h\to 0}\frac{g(x+h)-g(x)}{h} \] \[ = f^{\prime}(x) + g^{\prime}(x) \]

$\square$

Product rule of differentiation

Theorem

If \( f \) and \( g \) are differentiable at \( x \) then

\[ (fg)'(x) = f(x)g'(x) + f'(x)g(x) \]

Proof

As for the sum rule, we want to show that \( (fg)'(x) \) exists and is equal to \( f(x)g'(x) + f'(x)g(x) \). Recall that

\[ (fg)'(x) = \lim_{h \to 0} \frac{(fg)(x+h)-(fg)(x)}{h} \] \[ = \lim_{h \to 0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h} \] \[ = \lim_{h \to 0} \frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h} \] \[ = \lim_{h \to 0} \frac{f(x+h)g(x+h)-f(x+h)g(x)}{h} + \frac{f(x+h)g(x)-f(x)g(x)}{h} \] \[ = \lim_{h \to 0} f(x+h)\frac{g(x+h)-g(x)}{h} + \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}g(x) \] \[ = f(x)g'(x) + f'(x)g(x) \]

$\square$

Chain rule of differentiation

Theorem

If \( f \) and \( g \) are differentiable then

\[ (f \circ g)'(x) = f'(g(x))g'(x) \]

Proof

\[ (f \circ g)'(x) = \lim_{h \to 0} \frac{(f \circ g)(x + h) - (f \circ g)(x)}{h} \] \[ = \lim_{h \to 0} \frac{f(g(x + h)) - f(g(x))}{h} \] \[ = \lim_{h \to 0} \frac{f(g(x + h)) - f(g(x))}{g(x + h) - g(x)} \cdot \frac{g(x + h) - g(x)}{h} \] \[ = \left( \lim_{h \to 0} \frac{f(g(x + h)) - f(g(x))}{g(x + h) - g(x)} \right) \left( \lim_{h \to 0} \frac{g(x + h) - g(x)}{h} \right) \] \[ = f'(g(x))g'(x) \]

To justify this, we define \( H = g(x + h) - g(x) \), and equivalently \( g(x + h) = g(x) + H \). Note that

\[ \lim_{h \to 0} H = \lim_{h \to 0} g(x + h) - g(x) = g(x) - g(x) = 0 \]

By making the substitutions we have that

\[ \left( \lim_{h \to 0} \frac{f(g(x+h)) - f(g(x))}{g(x+h) - g(x)} \right) g'(x) = \left( \lim_{H \to 0} \frac{f(g(x) + H) - f(g(x))}{H} \right) g'(x) \] \[ = f'(g(x)) g'(x) \]

$\square$

Quotient rule of differentiation

Theorem

If \(f\) and \(g\) are differentiable at \(x\), then

\[ \left( \frac{f}{g} \right)'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g^2(x)} \]

Proof

Since \( f/g = f \cdot (1/g) \) we have

\[\left( \frac{f}{g} \right)' (a) = \left( f \cdot \frac{1}{g} \right)' (a)\] \[= f'(a) \cdot \left( \frac{1}{g} \right) (a) + f(a) \cdot \left( \frac{1}{g} \right)' (a)\] \[= \frac{f'(a)}{g(a)} + \frac{f(a)(-g'(a))}{[g(a)]^2}\] \[= \frac{f'(a) \cdot g(a) - f(a) \cdot g'(a)}{[g(a)]^2} \]

$\square$

Inverse functions

Definition

We say that \(g : D_g \rightarrow \mathbb{R}\) is the inverse function of \(f : D_f \rightarrow \mathbb{R}\) if for all \(x \in D_g\)

\[ (f \circ g)(x) = x = (g \circ f)(x) \]

and if for all \(x \in D_f\) we have

\[ (g \circ f)(x) = x \]

Remark

Note that in certain cases we might be interested in functions that are not defined over all of $\mathbb{R}$.

Differentiation of inverse functions

Proposition

If \( f^{-1} \) is differentiable at \( y \), then

\[ (f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))} \]

Proof

By the rule of composing functions, the derivative of \( f^{-1}(f(x)) \) is

\[ f'(x)(f^{-1})'(f(x)) \]

but because \( f^{-1}(f(x)) = x \) this is also equal to 1. Hence

\[ f'(x)(f^{-1})'(f(x)) = 1 \]

and so

\[ (f^{-1})'(f(x)) = \frac{1}{f'(x)} \]

Now setting \( f(x) = y \) we have \( x = f^{-1}(y) \) and so

\[ (f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))} \]

$\square$