Differentiability of Real Functions

Differentiation

Definition: Differentiability of a Real Functions

Let $f:\mathcal{D}\subseteq \mathbb{R}\to \mathbb{R}$ be a real function and let $S$ be some open subset of $\mathbb{R}$ which is contained in $\mathcal{D}$.

We say that $f$ is differentiable at $x_0\in S$ if the following limit exists:

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

In this case, the value of this limit is known as $f$‘s derivative at $x_0$.

NOTATION

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

If $f$ is differentiable at every $x\in S$, then we say that $f$ is differentiable on $S$ and if further $S=\mathcal{D}$, then we say that $f$ is just differentiable.

We also use the term “derivative” for the real function which to each $x\in S$ assigns the value of the above limit, if it exists.

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 the derivative $f^{\prime }$ and so on.

NOTATION

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

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).

Definition: Critical Point

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

We say that some $x_0\in \mathcal{D}$ is a critical point of $f$ if $f$ is not differentiable at $x_0$ or its derivative there is zero.

Theorem: Differentiability $⟹$ Continuity

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

PROOF

TODO

Mean Value Theorem for Derivatives

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

If $f$ is continuous on some 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

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

PROOF

TODO

Theorem: Darboux's Theorem

Let $f:\mathcal{D}\subseteq \mathbb{R}\to \mathbb{R}$ be a real function which 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 $f^{\prime }(a)<\lambda <f^{\prime }(b)$ or $f^{\prime }(a)>\lambda >f^{\prime }(b)$, there exists some $c\in (a;b)$ such that

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

PROOF

TODO

Theorem: Power Rule

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

$$(x^r{)}^{\prime }=r{x}^{r-1}$$

PROOF

TODO

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 at $x\in {\mathcal{D}}_f\cap {\mathcal{D}}_g$, then for all $\alpha ,\beta \in \mathbb{R}$ we have

$$(\alpha f(x)+\beta g(x){)}^{\prime }=\alpha f^{\prime }(x)+\beta g^{\prime }(x)$$

PROOF

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

Theorem: 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 at $x\in {\mathcal{D}}_f\cap {\mathcal{D}}_g$, then their product is also differentiable at $x$ with

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

PROOF

TODO

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 $g(x)\ne$, then their quotient $f/g$ is also differentiable at $x$ with

$${\left(\frac{f(x)}{g(x)}\right)}^{\prime }=\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{g(x{)}^2}$$

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

$$(f\circ g{)}^{\prime }(x)=f^{\prime }(g(x))g^{\prime }(x)$$

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 $f(x)>0$, then their exponentiation $f^g$ is also differentiable at $x$ with

$${\left(f(x{)}^{g(x)}\right)}^{\prime }=f(x{)}^{g(x)}\left(f^{\prime }(x)\frac{g(x)}{f(x)}+g^{\prime }(x)\ln (f(x))\right),$$

where $\ln$ is the real natural logarithm.

PROOF

TODO

Theorem: Derivative of the Inverse Function

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

If $f$ is differentiable at $x\in \mathcal{D}$ and $f^{\prime }(x)\ne 0$, then its inverse function $f^{-1}$ is differentiable at $y=f(x)$ with

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

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