Differentiability of Real Functions

Definition: Derivative of a Real Function

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

The derivative of $f$ at $x_0\in \mathcal{D}$ is the limit

$$\underset{\Delta x\to 0}{\lim }\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x},$$

if it exists.

NOTATION

$$f^{\prime }(x_0)\frac{\mathrm{d}f}{\mathrm{d}x}(x_0)\frac{\mathrm{d}}{\mathrm{d}x}f(x_0)$$

Note: Derivative Function

When we say just “the derivative of $f$”, we mean the function which maps every $x\in D$ to $f^{\prime }(x)$.

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 $f^{\prime }$;

• The second order derivative of $f$ is the derivative of $f^{\prime }$, etc.

NOTATION

$$f^{\prime }{f}^{\prime \prime }{f}^{\prime \prime \prime }{f}^{\mathrm{IV}}{f}^{\mathrm{V}}\cdots f^{(n)}$$ $$\frac{\mathrm{d}f}{\mathrm{d}x}\frac{{\mathrm{d}}^{2}f}{\mathrm{d}{x}^2}\frac{{\mathrm{d}}^{3}f}{\mathrm{d}{x}^3}\frac{{\mathrm{d}}^{4}f}{\mathrm{d}{x}^4}\frac{{\mathrm{d}}^{5}f}{\mathrm{d}{x}^5}\cdots \frac{{\mathrm{d}}^{n}f}{\mathrm{d}{x}^n}$$

Definition: Differentiability

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

We say that a real function $f$ is differentiable at some $x\in \mathcal{D}$ if the derivative of $f$ at $x$, i.e. $f^{\prime }(x)$, exists.

We say that $f$ is $k$-times (continuously) differentiable on $S\subseteq \mathcal{D}$ if its $k$-th order derivative exists (and is continuous).

Definition: Critical Point

We call $x_0\in \mathcal{D}$ a critical point of $f:\mathcal{D}\subseteq \mathbb{R}\to \mathbb{R}$ if $f$ is not differentiable at $x_0$ or its derivative at $x_0$ is zero.

Properties

Theorem: Differentiability $⟹$ Continuity

If $f:D\subseteq \mathbb{R}\in \mathbb{R}$ is differentiable at $x_0\in D$, then $f$ is also continuous at $x_0$.

PROOF

TODO

Mean Value Theorem for Derivatives

If $f:D\to \mathbb{R}$ is continuous on the closed interval $D=[a;b]\subset \mathbb{R}$ and differentiable on the open interval $(a;b)$, then there exists at least one $x_0\in (a;b)$ such that

$$f^{\prime }(x_0)=\frac{f(b)-f(a)}{b-a}$$

PROOF

TODO

Intuition: Geometric Meaning

The theorem says that there is at least one point $(x_0;f(x_0))$ on the graph of $f$, where the tangent line to $f$ is parallel to the secant line through the points $(a;f(a))$ and $(b;f(b))$.

TODO

Theorem: Darboux's Theorem

Let $f:D\to \mathbb{R}$ be a differentiable real function on a closed interval $D=[a;b]\subset \mathbb{R}$.

For each $\lambda \in \mathbb{R}$ such that $f^{\prime }(a)<\lambda <f^{\prime }(b)$ or $f^{\prime }(a)>\lambda >f^{\prime }(b)$, there exists some $c\in (a;b)$ such that

$$f^{\prime }(c)=\lambda$$

PROOF

TODO

Differentiation Rules

Theorem: Arithmetic Differentiation Rules

If $f,g:D\subseteq \mathbb{R}\to \mathbb{R}$ are differentiable at $x\in D$, then the following hold:

• Linearity: $[\alpha f(x)+\beta g(x){]}^{\prime }=\alpha f^{\prime }(x)+\beta g^{\prime }(x)$ for all $\alpha ,\beta \in \mathbb{R}$

• Product Rule: $[f(x)g(x){]}^{\prime }=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$

• Quotient Rule: ${\left[\frac{f(x)}{g(x)}\right]}^{\prime }=\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{g(x{)}^2}$ provided that $g(x)\ne 0$

• General Power Rule: ${\left[f(x{)}^{g(x)}\right]}^{\prime }=f(x{)}^{g(x)}\left[f^{\prime }(x)\frac{g(x)}{f(x)}+g^{\prime }(x)\ln (f(x))\right]$

PROOF

Proof of linearity:

$$\begin{array}{rl}[\lambda f(x)+\mu g(x){]}^{\prime }& =\underset{\Delta x\to 0}{\lim }\frac{\lambda f(x_0+\Delta x)+\mu g(x_0+\Delta x)-\lambda f(x_0)-\mu g(x_0)}{\Delta x}\\ & =\underset{\Delta x\to 0}{\lim }\frac{\lambda [f(x_0+\Delta x)-f(x_0)]+\mu [g(x_0+\Delta x)-g(x_0)]}{\Delta x}\\ & =\underset{\Delta x\to 0}{\lim }\lambda \frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}+\underset{\Delta x\to 0}{\lim }\mu \frac{g(x_0+\Delta x)-g(x_0)}{\Delta x}\\ & =\lambda \underset{\Delta x\to 0}{\lim }\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}+\mu \underset{\Delta x\to 0}{\lim }\frac{g(x_0+\Delta x)-g(x_0)}{\Delta x}\\ & =\lambda f^{\prime }(x_0)+\mu g^{\prime }(x_0)\end{array}$$

Theorem: The Chain Rule

If $g:D\subseteq \mathbb{R}\to \mathbb{R}$ is differentiable at $x\in D$ and $f:g(D)\to \mathbb{R}$ is differentiable at $g(x)$, then

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

PROOF

TODO

Theorem: Derivative of the Inverse Function

Let $f:D\subseteq \mathbb{R}\to \mathbb{R}$ be invertible.

If $f$ is differentiable at $f^{-1}(y),y\in f(D)$ and $f^{\prime }(f^{-1}(y))\ne 0$, then

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

PROOF

TODO

Common Derivatives

Theorem: Power Rule

$$(x^{\alpha }{)}^{\prime }=\alpha x^{\alpha -1}\forall \alpha \in \mathbb{R}$$

PROOF

TODO

Theorem: Derivative of Exponential Functions

$$({\mathrm{e}}^x{)}^{\prime }={\mathrm{e}}^x$$ $$(a^x{)}^{\prime }=a^x\ln a$$

PROOF

$$\begin{array}{rl}(e^x{)}^{\prime }& =\underset{\Delta x\to 0}{\lim }\frac{e^{x+\Delta x}-e^x}{\Delta x}=\underset{\Delta x\to 0}{\lim }\frac{e^x(e^{\Delta x}-1)}{\Delta x}=e^x\underset{\Delta x\to 0}{\lim }\frac{e^{\Delta x}-1}{\Delta x}\\ & =e^x\underset{\Delta x\to 0}{\lim }\frac{\underset{n\to \infty }{\lim }{\left(1+\frac{\Delta x}{n}\right)}^n-1}{\Delta x}=e^x\underset{\Delta x\to 0}{\lim }\underset{n\to \infty }{\lim }\frac{(1+\frac{\Delta x}{n}{)}^n-1}{\Delta x}\\ & =e^x\underset{\Delta x\to 0}{\lim }\underset{n\to \infty }{\lim }\frac{1+n\frac{\Delta x}{n}+\left(\begin{array}{c}n\\ 2\end{array}\right){\left(\frac{\Delta x}{n}\right)}^2+\cdots +\left(\begin{array}{c}n\\ n-1\end{array}\right){\left(\frac{\Delta x}{n}\right)}^{n-1}+{\left(\frac{\Delta x}{n}\right)}^n-1}{\Delta x}\\ & =e^x\underset{\Delta x\to 0}{\lim }\underset{n\to \infty }{\lim }\left(1+\left(\begin{array}{c}n\\ 2\end{array}\right)\frac{\Delta x}{n^2}+\cdots +\left(\begin{array}{c}n\\ n-1\end{array}\right)\frac{\Delta x^{n-2}}{n^{n-1}}+\frac{\Delta x^{n-1}}{n^n}\right)\\ & =e^x\underset{n\to \infty }{\lim }\underset{\Delta x\to 0}{\lim }\left(1+\underset{\to 0}{\underbrace{\left(\begin{array}{c}n\\ 2\end{array}\right)\frac{\Delta x}{n^2}+\cdots +\left(\begin{array}{c}n\\ n-1\end{array}\right)\frac{\Delta x^{n-2}}{n^{n-1}}+\frac{\Delta x^{n-1}}{n^n}}}\right)\\ & =e^x\underset{n\to \infty }{\lim }1=e^x\end{array}$$ $$\begin{array}{r}(a^x{)}^{\prime }=((e^{\ln a}{)}^x{)}^{\prime }=(e^{x\ln a}{)}^{\prime }=e^{x\ln a}\ln a=(e^{\ln a}{)}^x\ln a=a^x\ln a\end{array}$$

Theorem: Derivative of Logarithmic Functions

$$(\ln x{)}^{\prime }=\frac{1}{x}$$ $$({\log }_{a}x{)}^{\prime }=\frac{1}{x\ln a}$$

PROOF

TODO