Differentiability (Real Functions)#

Definition: Differentiability (Real Functions)

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function and let \(p \in \mathcal{D}\) be a limit point of \(\mathcal{D}\).

We say that \(f\) is differentiable at \(p\) if the limit

\[\lim_{x \to p} \frac{f(x) - f(p)}{x - p}\]

exists and is finite. In this case, the value of this limit is known as \(f\)'s derivative at \(p\).

Notation

We denote \(f\)'s derivative at \(p\) as \(f'(p)\). If a specific label such as \(x\), \(t\), etc. is used for \(f\)'s input, then we also denote it in one of the following ways:

\[\left.\frac{\mathrm{d}f}{\mathrm{d}x}\right\vert_{x=p} \qquad \left.\frac{\mathrm{d}f}{\mathrm{d}t}\right\vert_{t=p}\]

Theorem: Differentiability at Interior Points

A real function \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) is differentiable at an interior point \(p\) of \(\mathcal{D}\) if and only if the limit

\[\lim_{h \to 0} \frac{f(p+h) - f(p)}{h}\]

exists and is finite. In this case, the value of this limit is \(f'(p)\).

Example: \(f(x) = ax + b\)

We want to find the derivative at each \(x \in \mathbb{R}\) of the function \(f: \mathbb{R} \to \mathbb{R}\) defined as

\[f(x) = ax + b,\]

where \(a, b \in \mathbb{R}\). Since every \(x\) is an interior point of \(\mathbb{R}\), we have:

\[\begin{aligned}f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} &= \frac{a(x+h) + b - (ax + b)}{h} \\ &= \lim_{h\to 0}\frac{ax + ah + b - ax - b}{h} \\ &= \lim_{h \to \infty} \frac{ah}{h} = a\end{aligned}\]

Therefore:

\[f'(x) = a\]

Example: \(f(x) = x^2\)

We want to find the derivative at each \(x \in \mathbb{R}\) of the following function \(f: \mathbb{R} \to \mathbb{R}\):

\[f(x) = x^2\]

Since every \(x\) is an interior point of \(\mathbb{R}\), we have:

\[\begin{aligned}f'(x) = \lim_{h \to 0} \frac{(x + h)^2 - x^2}{h} &= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 -x^2}{h} \\ & = \lim_{h \to 0} \frac{2xh + h^2}{h} \\ &= \lim_{h \to 0} (2x + h) = 2x\end{aligned}\]

Proof

TODO

Definition: Critical Point

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function and let \(p \in \mathcal{D}\).

We say \(p\) is a critical point of \(f\) if \(f\) is not differentiable at \(x_0\) or its derivative there is zero.

Theorem: Differentiability \(\implies\) Continuity

If \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) is differentiable at \(p \in \mathcal{D}\), then \(f\) is also continuous at \(p\).

Proof

We need to show that \(f(p)\) is equal to the [limit](./Limits%20(Real%20Functions.md) of \(f\) at \(p\):

\[ \lim_{x \to p} f(x) = f(p) \]

With the substition \(h = x - p\), we get the following:

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

This means that we need to show that \(\lim_{h \to 0} (f(p + h) - f(p)) = 0\).

We have the following:

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

Mean Value Theorem for Derivatives

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function.

If \(f\) is continuous on the closed interval \([a;b]\) and is differentiable on the open interval \((a;b)\), then there exists as least one \(\xi \in (a;b)\) such that

\[ f'(\xi) = \frac{f(b) - f(a)}{b - a} \]

Proof

TODO

Theorem: Darboux's Theorem

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function and let \([a;b]\) be a closed interval such that \([a;b] \subseteq \mathcal{D}\).

If \(f\) is differentiable on the open interval \((a;b)\), then for each \(\lambda \in \mathbb{R}\) such that \(f'(a) \lt \lambda \lt f'(b)\) or \(f'(a) \gt \lambda \gt f'(b)\), there exists some \(c \in (a;b)\) such that

\[ f'(c) = \lambda \]

Proof

TODO

Theorem: Power Rule

For all \(r \in \mathbb{R}\), the exponentiation \(x^r\) is differentiable with

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

Proof

TODO

Example: \((\sqrt{x})'\)

We can use this to find the derivative of \(\sqrt{x}\):

\[ (\sqrt{x})' = \left(x^{\frac{1}{2}}\right)' = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}} \]

Theorem: Linearity of Differentiation

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) and \(g: \mathcal{D}_g \subseteq \mathbb{R} \to \mathbb{R}\) be a real functions.

If \(f\) and \(g\) are both differentiable on \(S \subseteq \mathcal{D}_f \cap \mathcal{D}_g\), then \(\alpha f + \beta g\) is also differentiable on \(S\) for all \(\alpha, \beta \in \mathbb{R}\) with

\[ (\alpha f + \beta g)' = \alpha f' + \beta g' \]

Proof

\[ \begin{aligned}[\lambda\, f(x) + \mu\, g(x)]' &= \operatorname*{lim}_{h\rightarrow0}\frac{\lambda f(x_{0}+h)+\mu g(x_{0}+h) - \lambda f(x_{0})-\mu g(x_{0})}{h}\\ &= \lim_{h\to 0}\frac{\lambda[f(x_{0}+h)-f(x_{0})] + \mu[g(x_{0}+h)-g(x_{0})]}{h} \\ &= \lim_{h\to 0}\lambda\frac{f(x_0+h)-f(x_0)}{h} + \lim_{h\to 0}\mu\frac{g(x_0+h)-g(x_0)}{h} \\ &= \lambda\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h} + \mu\lim_{h\to 0}\frac{g(x_0+h)-g(x_0)}{h} \\ &= \lambda f'(x_0) + \mu g'(x_0)\end{aligned} \]

Theorem: Product Rule

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) and \(g: \mathcal{D}_g \subseteq \mathbb{R} \to \mathbb{R}\) be a real functions.

If \(f\) and \(g\) are both differentiable on \(S \subseteq \mathcal{D}_f \cap \mathcal{D}_g\), then their product \(fg\) is also differentiable on \(S\) with

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

Proof

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

Since \(f\) is differentiable at \(x\), it must also be continuous there, i.e. \(\lim_{h \to 0}f(x +h) = f(x)\). Therefore, we have:

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

Theorem: Quotient Rule

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) and \(g: \mathcal{D}_g \subseteq \mathbb{R} \to \mathbb{R}\) be a real functions.

If \(f\) and \(g\) are both differentiable at \(x \in \mathcal{D}_f \cap \mathcal{D}_g\) and \(g(x) \ne 0\), then their quotient \(f/g\) is also differentiable at \(x\) with

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

Proof

TODO

Theorem: Chain Rule

Let \(g: \mathcal{D}_g \subseteq \mathbb{R} \to \mathbb{R}\) be a real function and let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) also be a real function such that the image of \(g\) is contained in the domain of \(f\), i.e. \(g(\mathcal{D}_g) \subseteq \mathcal{D}_f\).

If \(g\) is differentiable at some \(x \in \mathcal{D}_g\) and \(f\) is differentiable at \(g(x)\), then their composition \(f \circ g: \mathcal{D}_g \to \mathbb{R}\) is also differentiable at \(x\) with

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

Proof

TODO

Theorem: General Power Rule

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) and \(g: \mathcal{D}_g \subseteq \mathbb{R} \to \mathbb{R}\) be a real functions.

If \(f\) and \(g\) are both differentiable at \(x \in \mathcal{D}_f \cap \mathcal{D}_g\) and \(f(x) \gt 0\), then the exponentiation \(f^g\) is also differentiable at \(x\) with

\[ \left(f(x)^{g(x)}\right)' = f(x)^{g(x)}\left(f'(x)\frac{g(x)}{f(x)} + g'(x)\ln(f(x))\right), \]

where \(\ln\) is the real natural logarithm.

Proof

TODO

Theorem: Derivatives of Inverse Functions

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be an injective real function and let \(y \in f(\mathcal{D})\).

If \(f\) is differentiable at \(f^{-1}(y)\) with \(f'(f^{-1}(y)) \ne 0\) and its inverse function \(f^{-1}: f(\mathcal{D}) \to \mathcal{D}\) is continuous at \(y\), then \(f^{-1}\) is differentiable at \(y\) with

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

Proof

Since \(y \in f(\mathcal{D})\), we have \(y = f(x)\) for some unique \(x \in \mathcal{D}\).

TODO

Example: \((\arcsin x)'\)

The restriction of the sine function on \(\left[\frac{-\pi}{2}; \frac{\pi}{2}\right]\) is injective and differentiable with \(\sin '(x) \ne 0\). We can therefore use the aforementioned theorem to find the derivative of the arcsine function:

\[ (\arcsin y)' = \frac{1}{\sin'(\arcsin y)} = \frac{1}{\cos (\arcsin y)} \]

On \(\left[\frac{-\pi}{2}; \frac{\pi}{2}\right]\) we know that \(\cos x = \sqrt{1 - \sin x}\) and since \(\sin(\arcsin y) = y\), we get:

\[ (\arcsin y)' = \frac{1}{\sqrt{1-y^2}} \]

Higher-Order Differentiability#

Definition: Higher-Order Derivatives

The \(k\)-th order derivative of \(f\) is the derivative of the \((k-1)\)-th order derivative of \(f\):
- The \(0\)-th order derivative of \(f\) is just \(f\) itself;
- The first order derivative of \(f\) is just the aforementioned derivative \(f'\);
- The second order derivative of \(f\) is the derivative of the derivative \(f'\) and so on.

Notation

\[ f' \qquad f'' \qquad f''' \qquad f^{\mathrm{IV}} \qquad f^{\mathrm{V}} \qquad \cdots \qquad f^{(n)} \]

\[ \frac{\mathrm{d}f}{\mathrm{d}x} \qquad \frac{\mathrm{d}^2 f}{\mathrm{d}x^2} \qquad \frac{\mathrm{d}^3 f}{\mathrm{d}x^3} \qquad \frac{\mathrm{d}^4 f}{\mathrm{d}x^4} \qquad \frac{\mathrm{d}^5 f}{\mathrm{d}x^5} \qquad \cdots \qquad \frac{\mathrm{d}^n f}{\mathrm{d}x^n} \]

Example: \(f(x) = x^3 - 7x^2 + 2x -3\)

For the function \(f(x) = x^3 - 7x^2 + 2x -3\) we have the following:

\[ \begin{aligned}f'(x) &= 3x^2 - 14x +2 \\ f''(x) &= 6x-14 \\ f'''(x) &= 6 \\ f^{(4)}(x) &= 0\end{aligned} \]

If \(f\)'s \(k\)-order derivative exists (and is continuous), then we say that \(f\) is \(k\)-times (continuously) differentiable (at some point or on some subset).

Notation

If \(f\) is \(k\)-times continuously differentiable on \(S\), we write \(f \in C^n(S)\).

Theorem: Higher-Order Product Rule

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) and \(g: \mathcal{D}_g \subseteq \mathbb{R} \to \mathbb{R}\) be a real functions.

If \(f\) and \(g\) are both \(n\)-times differentiable on \(S \subseteq \mathcal{D}_f \cap \mathcal{D}_g\), then their product is also \(n\)-times differentiable on \(S\) with

\[(fg)^{(n)} = \sum_{k = 0}^n \binom{n}{k} f^{(n-k)}g^{(k)}\]

Proof

The proof is by induction.

Base case (\(n = 0\)):

We have \((fg)^{(0)} = fg\) and \(\sum_{k = 0}^0 \binom{n}{k} f^{(n-k)}g^{(k)} = f^{(0 - 0)}g^{(0)} = fg\).

Induction hypothesis: There exists some \(n \in \mathbb{N}_0\) such that \((fg)^{(n)} = \sum_{k = 0}^n \binom{n}{k} f^{(n-k)}g^{(k)}\) implies that \((fg)^{(n+1)} = \sum_{k = 0}^{n+1} \binom{n+1}{k} f^{(n-k+1)}g^{(k)}\).

Inductive step:

\[\begin{aligned}(fg)^{(n+1)} = ((fg)^{(n)})' & = \left(\sum_{k = 0}^n \binom{n}{k} f^{(n-k)}g^{(k)}\right)' \\ & = \sum_{k = 0}^n \binom{n}{k}(f^{(n-k)}g^{(k)})' \\ & = \sum_{k = 0}^n \binom{n}{k} (f^{(n-k)})'g^{(k)} + f^{(n-k)}(g^{(k)})'\\ & = \sum_{k = 0}^{n+1} \binom{n+1}{k} f^{(n-k+1)}g^{(k)}\end{aligned}\]