Differentiability
How to quantify the rate of change of a function.
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Study Differentiability
To motivate the problem, consider the functions \(f,g:\mathbb{R}\to\mathbb{R}\) with \(f(x)=x^2\) and \(g(x)=|x|\).
As we already know, both these functions are continuous. Let us now focus on the qualitative behavior of the functions at the origin.
| \(f\) | \(g\) |
|---|---|
| smooth | sharp bend |
| there is a unique tangent | no unique tangent |
A straight line \(y(t)\) going through the points \((x_0,f(x_0))\) and \((x,f(x))\) is called secant 1 of \(f\) through these points. It is given by \[y(t)=f(x_0)+\frac{f(x_0)-f(x)}{x_0-x}(t-x_0) .\] In particular, the slope of \(y\) is the difference quotient \[\frac{f(x_0)-f(x)}{x_0-x}=\frac{f(x)-f(x_0)}{x-x_0}.\]
If we now let \(x\) tend to \(x_0\), we obtain a tangent of \(f\) at \(x_0\). This leads to the following definition.
Definition 1. Let \(I\subset\mathbb{R}\) be an interval with
more than one point or an open set. Let \(f:I\to\mathbb{R}\) be a function. Then
\(f\) is called differentiable at
\(x_0\in I\) if there exists
a function \(\Delta_{f,x_0}:I\to\mathbb{R}\) that is
continuous in \(x_0\) and, moreover,
for all \(x\in I\) holds \[f(x)=f(x_0)+(x-x_0)\cdot\Delta_{f,x_0}(x).\]
The number \(\Delta_{f,x_0}(x_0)\) is
called derivative of \(f\) at \(x_0\).
The function \(f\) is called
differentiable in \(I\) if it
is differentiable at all \(x_0\in
I\).
By solving the above equation for \(\Delta_{f,x_0}(x)\), we get for \(x\neq x_0\) that \[\Delta_{f,x_0}(x)=\frac{f(x)-f(x_0)}{x-x_0},\]
i.e., it is the difference quotient. Continuity of \(\Delta_{f,x_0}(x)\) at \(x_0\) is therefore equivalent to the
existence of the limit \[\lim_{x\to
x_0}\Delta_{f,x_0}(x)=\lim_{x\to
x_0}\frac{f(x)-f(x_0)}{x-x_0}=:f'(x_0).\] Also the following
notation is used in the literature for \(f'(x_0)\):
\(\frac{d}{dx}f(x_0)\), \(\frac{\partial}{\partial x}f(x_0)\), \(\frac{d f}{d x}|_{x=x_0}\), \(\frac{\partial f}{\partial x}|_{x=x_0}\),
\(\partial_xf(x_0)\).
The next result states that differentiability is a stronger property
than continuity.
Theorem 2. Let \(f:I\to\mathbb{R}\) be differentiable at \(x_0\in I\). Then \(f\) is continuous in \(x_0\).
Proof: By writing \[f(x)=f(x_0)+(x-x_0)\cdot\Delta_{f,x_0}(x),\] the continuity of \(\Delta_{f,x_0}\) at \(x_0\) implies the continuity of \(f\) at \(x_0\).\(\Box\)
As the following example shows, the opposite implication cannot be made, i.e., not every continuous function is differentiable.
Example 3. Consider the absolute value function
\(|\cdot|:\mathbb{R}\to\mathbb{R}\). We
already know that it is continuous. For the analysis of
differentiability, we distinguish between three cases: 1st
Case: \(x_0>0\).
Then we have that \(|x_0|=x_0\) and,
moreover, for \(x\) in some
neighbourhood of \(x_0\) holds \(|x|=x\). Therefore, we have \[\lim_{x\to x_0}\frac{|x|-|x_0|}{x-x_0}=\lim_{x\to
x_0}\frac{x-x_0}{x-x_0}=1.\] 2nd Case:
\(x_0<0\).
Then we have that \(|x_0|=-x_0\) and,
moreover, for \(x\) in some
neighbourhood of \(x_0\) holds \(|x|=-x\). Therefore, we have \[\lim_{x\to x_0}\frac{|x|-|x_0|}{x-x_0}=\lim_{x\to
x_0}\frac{-x+x_0}{x-x_0}=-1.\] 3rd Case:
\(x_0=0\).
Then the two sequences \((x_n)_{n\in\mathbb{N}}=(\frac1n)_{n\in\mathbb{N}}\),
\((y_n)_{n\in\mathbb{N}}=(-\frac1n)_{n\in\mathbb{N}}\)
both tend to \(x_0=0\). However, we
have \[\lim_{n\to
\infty}\frac{|x_n|-|x_0|}{x_n-x_0}=
\lim_{n\to \infty}\frac{|\frac1n|-|0|}{\frac1n-0}=1\] and \[\lim_{n\to \infty}\frac{|y_n|-|x_0|}{y_n-x_0}=
\lim_{n\to \infty}\frac{|-\frac1n|-|0|}{-\frac1n-0}=-1.\]
Therefore, the limit \[\lim_{x\to
0}\frac{|x|-|0|}{x-0}\] does not exist. Thus, \(|\cdot|\) is not differentiable at \(x_0=0\).
Note that, by a substitution \(h:=x-x_0\), the difference quotient can be reformulated as \[f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}.\]
The derivative \(a:=f'(x_0)\) of \(f\) in \(x_0\) can be interpreted in the following way: The linear mapping \(\varphi:\mathbb{R}\rightarrow\mathbb{R},\ x\mapsto ax\) fulfills \[\lim_{h\rightarrow 0}\frac{|f(x_0 + h)-(f(x_0)+\varphi(h))|}{|h|}=\lim_{h\rightarrow 0}\left|\frac{f(x_0+h)-f(x_0)}{h} - a \right| = 0.\] This means that the affine linear mapping \(t(h):=f(x_0)+\varphi(h)=f(x_0)+ah\), which actually is the tangent of \(f\) at \(x_0\), approximates \(f(x)\) linearly in a neighbourhood of \(x_0\) in a best possible way.
Now we consider some examples of differentiable functions.
Example 4.
Given is some constant \(c\in\mathbb{R}\). Consider the constant function \(f:\mathbb{R}\to\mathbb{R}\) with \(f(x)=c\) for all \(x\in \mathbb{R}\). Then for all \(x_0\in \mathbb{R}\) holds \[f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}\frac{c-c}{x-x_0}=0.\]
Given is some constant \(c\in\mathbb{R}\). Consider the linear function \(f:\mathbb{R}\to\mathbb{R}\) with \(f(x)=cx\) for all \(x\in \mathbb{R}\). Then for all \(x_0\in \mathbb{R}\) holds \[f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}\frac{cx-cx_0}{x-x_0}=\lim_{x\to x_0}c\frac{x-x_0}{x-x_0}=c.\]
For determining the derivatives of \(\exp\), \(\sinh\), \(\cosh\), \(\sin\) and \(\cos\), we first determine the following limit for \(\lambda\in\mathbb{C}\): \[\lim_{h\to0}\frac{\exp(\lambda h)-1}{h}.\] By Theorem [thm:expest], we know that for \(h\in\mathbb{R}\) with \(|\lambda h|<2\) \[\exp(\lambda h)=1+\lambda h+r_2(\lambda h)\] with \(|r_2(\lambda h)|\leq |\lambda h|^2\). Therefore, \[\lim_{h\to0}\frac{\exp(\lambda h)-1}{h}=\lim_{h\to0}\frac{1+\lambda h+r_2(\lambda h)-1}{h}=\lim_{h\to0}\left(\lambda+\frac{r_2(\lambda h)}{h}\right) =\lambda.\] We can further conclude \[\lim_{h\to0}\frac{\exp(\lambda(x_0+h))-\exp(\lambda x_0)}{h}= \exp(\lambda x_0)\cdot\lim_{h\to0}\frac{\exp(\lambda h)-1}{h}=\lambda \exp(\lambda x_0).\] This has manifold consequences for the derivatives of exponential, hyperbolic and trigonometric functions: \[\exp'(x_0)=\lim_{h\to0}\frac{\exp(x_0+h)-\exp(x_0)}{h}=\exp(x_0),\] i.e., \(\exp'=\exp\).
We can further conclude that \[\begin{aligned} \sinh'(x_0)=&\lim_{h\to0}\frac{\sinh(x_0+h)-\sinh(x_0)}{h}\\=& \frac12\lim_{h\to0}\left(\frac{\exp(x_0+h)-\exp(x_0)}{h}+\frac{-\exp(-(x_0+h))+\exp(-x_0)}{h}\right)\\ =&\frac12(\exp(x_0)+\exp(-x_0))=\cosh(x_0). \end{aligned}\] Analogously, we can show that \(\cosh'=\sinh\). Now consider the trigonometric functions: \[\begin{aligned} \sin'(x_0)=&\lim_{h\to0}\frac{\sin(x_0+h)-\sin(x_0)}{h}\\ =&\frac1{2i}\lim_{h\to0}\left(\frac{\exp(i(x_0+h))-\exp(ix_0)}{h}+\frac{-\exp(-i(x_0+h))+\exp(-ix_0)}{h}\right)\\ =&\frac1{2i}(i\exp(ix_0)+i\exp(-ix_0))=\cos(x_0) \end{aligned}\] and \[\begin{aligned} \cos'(x_0)=&\lim_{h\to0}\frac{\cos(x_0+h)-\cos(x_0)}{h}\\ =&\frac1{2}\lim_{h\to0}\left(\frac{\exp(i(x_0+h))-\exp(ix_0)}{h}+\frac{\exp(-i(x_0+h))-\exp(-ix_0)}{h}\right)\\ =&\frac1{2}(i\exp(ix_0)-i\exp(-ix_0))\\=& -\frac1{2i}(\exp(ix_0)-\exp(-ix_0))=-\sin(x_0). \end{aligned}\]
from the Latin word secare = “to cut”↩︎
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