Derivatives

Limits §

In this section, we discuss limits and what they mean.

A point $x_0\in \mathbb{R}$ is a limit point of $D\subseteq \mathbb{R}$, provided there exists some sequence $\{x_n\}$ in $D/\{x_0\}$ such that

$$\{x_n\}\to x_0$$

Example: Limit Points of $\mathbb{R}$

What are limit points of $D=\mathbb{R}$?

All of the points in $\mathbb{R}$, as for any $x_0$, we can define the sequence

$$\{x_n\}=x_0+\frac{1}{n}$$

Which wil converge to $x_0$.

Example: Limit Points of Intervals

Let $D=(0,1)$. What are the limit points?

All of the points in the interval $[0,1]$, as we can define a sequence that converges to the endpoints, or any value inside of the interval.

Let $f:D\to \mathbb{R}$ be a function, and let $x_0\in \mathbb{R}$ be a limit point of $D$. Then, the limit of $f$ is

$$\underset{x\to x_0}{\lim }f(x)=L$$

If for any sequence $\{x_n\}$ in $D/\{x_0\}$, if $\{x_n\}\to x_0$, then $f(x_n)\to L$.

Limits can only be defined if there exists a sequence in our domain that converges to the point (by definition of a limit point).

Formally, for all $ϵ>0$, there exists a $\delta >0$ such that for all $x\in D$,

$$0<|x-x_0|<\delta \to |f(x)-L|<ϵ$$

Note that all of our limit rules apply, and can be used.

Derivatives §

Definition §

Consider a function $f:I\to \mathbb{R}$ defined along some interval. Say we want the “slope” of the function at some limit point $x_0\in I$.

This is the derivative of the function, $f^{\prime }(x_0)$, defined as

$$f^{\prime }(x_0)=\underset{x\to x_0}{\lim }\frac{f(x)-f(x_0)}{x-x_0}=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x_0)}{h}$$

Where $\{x\}\in I/\{x_0\}$ is any sequence converging to $x_0$.

Note you can obtain the decond definition from the first by setting $x=x_0+h$

If the derivative for $f$ exists at a point $x_0$, we say $f$ is differentiable at this point.

Theorem: Differentiability and Continuity

If $f^{\prime }(x_0)$ exists, then $f$ is continuous at $x_0$. In other words, differentiability implies continuity.

Proof

Let $x_n\to x_0$, where $x_n\in I/\{x_n\}$. We want to show that $f(x_n)\to f(x_0)$.

$$\begin{array}{r}f(x_n)-f(x_0)=\frac{f(x_n)-f(x_0)}{x_n-x_0}(x_n-x_0)\to f^{\prime }(x_0)\cdot 0=0\end{array}$$

Derivatives have a variety of useful properties that we can prove! They are given below.

Theorem: Product Rule

If $f:I\to \mathbb{R}$ and $g:I\to \mathbb{R}$ are both differentiable at $x_0$, so is $f\cdot g$.

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

Proof

Let $x_n\to x_0$ with $x_n\in I/\{x_0\}$. Then,

$$\begin{array}{rl}\underset{x_n\to x_0}{\lim }\frac{f(x_n)g(x_n)-f(x_0)g(x_0)}{x_n-x_0}& =\underset{x_n\to x_0}{\lim }\frac{f(x_n)g(x_n)-f(x_0)g(x_0)+f(x_n)g(x_0)-f(x_n)g(x_0)}{x_n-x_0}\\ & =\underset{x_n\to x_0}{\lim }\frac{f(x_n)(g(x_n)-g(x_0))+g(x_0)(f(x_n)-f(x_0))}{x_n-x_0}\\ & =\underset{x_n\to x_0}{\lim }f(x_n)\frac{g(x_n)-g(x_0)}{x_n-x_0}+g(x_0)\frac{f(x_n)-f(x_0)}{x_n-x_0}\\ & =f(x_0)g^{\prime }(x_0)+g(x_0)f^{\prime }(x_0)\end{array}$$

Theorem: Chain Rule

If $f:I\to \mathbb{R}$ and $g:I\to \mathbb{R}$ are both differentiable at $x_0$, so is $f(g(x))$.

$$\frac{d}{dx}f(g(x))=f^{\prime }(g(x))g^{\prime }(x)$$

Theorem: Derivatives of Inverses

Let $x_0\in I$ and $f:I\to \mathbb{R}$ be a strictly monotone continuous function. Suppose $f^{\prime }(x_0)\ne 0$ exists (to avoid division by 0). Then, we can find the derivative of $f$’s inverse as follows:

$$(f^{-1}{)}^{\prime }(x_0)=\frac{1}{f^{\prime }(y_0)}=\frac{1}{f^{\prime }(f^{-1}(x_0))}$$

Example: Derivatives of Inverses

Let $f(x)=x^2$, where $f:[0,\infty )\to \mathbb{R}$. Find $(f^{-1}{)}^{\prime }(4)$.

We have $f^{\prime }(x)=2x$, and $f^{-1}(4)=2$. Then,

$$(f^{-1}{)}^{\prime }(4)=\frac{1}{2(2)}=\frac{1}{4}$$

Mean-Value Theorem §

Definition and Proof §

Let $f$ be a function continuous on $[a,b]$, and differentiable on $(a,b)$. Then $\exists c\in (a,b)$ such that

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

In other words, there exists a point where the derivative is equal to the average slope!

Example: Mean-Value Theorem Applications

$\forall x$, let $|f^{\prime }(x)|\le M$. Then, by mean value theorem,

$$|f(x)-f(y)|=|f^{\prime }(c)(x-y)|\le M|x-y|$$

This proves an earlier theorem in uniform continuity!

We will prove this theorem; but first, we define a variety of lemmas that will be useful to us.

Lemma

Let $I$ be an open interval containing $x_0$, and $f:I\to \mathbb{R}$. Suppose $f^{\prime }(x_0)$ exists,

If $f(x_0)$ is a max (or min), then $f^{\prime }(x_0)=0$.

Proof

Without loss of generality, assume $f(x_0)$ is a max. We show that the limit of slopes on the left and right bound $f^{\prime }(x_0)$, forcing $f^{\prime }(x_0)=0$.

Define the interval $I=(x_0-1/n,x_0+1/n)$ sufficiently close to our maximum. We have

$$\frac{f\left(x_0+\frac{1}{n}\right)-f(x_0)}{\left(x+\frac{1}{n}\right)-x_0}\le 0$$

Take this from $n\to \infty$ to get $f^{\prime }(x_0)\le 0$.

Similarly, we define the sequence $x_n=x_0-1/n$ to show that $f^{\prime }(x_0)\ge 0$, which forces $f^{\prime }(x_0)$ as $0\le f^{\prime }(x_0)\le 0$.

Theorem: Rolle's Lemma

Let $f$ be continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$. Then, $\exists c\in (a,b)$ such that $f^{\prime }(c)=0$.

This is a special case of MVT, but we will use it to prove MVT!

Proof

By the extreme value theorem, we know that the maximum and minimum of $f$ exist. We do a case analysis:

1. Suppose the max and min both occur at the endpoints. Then, the function is constant - we can choose any point in the interior whose derivative is 0.
2. Suppose a max (or min) appears in $(a,b)$. Then, we apply our previous lemma to guarantee that $\exists c\in (a,b)$ such that $f^{\prime }(c)=0$.

We now prove MVT. The idea of this proof is to create a function from $f$ whose endpoints are the same, and applying Rolle’s theorem.

Proof (Mean Value Theorem)

Let $f$ be a function continuous on $[a,b]$, and differentiable on $(a,b)$. Let $h(x)=f(x)-mx$ for some $m\in \mathbb{R}$. We can choose an $m$ such that $h(a)=h(b)$ in order to apply Rolle’s theorem.

$$f(a)+ma=f(b)+mb⟹m=\frac{f(b)-f(a)}{b-a}$$

By Rolle’s lemma, $\exists c\in (a,b)$ such that $h^{\prime }(c)=0$! Plugging this in and solving for $f^{\prime }(c)$, this gives us

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

Example: Mean Value Theorem

Prove that $5x+\sin (x)=0$ does not have more than one solution.

Note that $f^{\prime }(x)=5+\cos (x)\le 4>0$. If there were two or more solutions, by Rolle’s lemme, $f^{\prime }(x)=0$ at some point, which is a contradiction.

Corollary 1

If $f:I\to \mathbb{R}$ is differentiable, then $f^{\prime }(x)=0$ for all $x$ is true if and only if $f$ is constant.

Proof

Proof ($\leftarrow$) §

We can use limit definition to easily show that the derivative is 0.

Proof ($\to$) §

By contrapositive, suppose that $f$ is not constant. Then there exists two separate points $a,b\in I$ such that $f(a)\ne f(b)$.

By MVT, there exists a derivative such that

$$f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}\ne 0$$

We apply the contrapositive to obtain our corollary.

Corollary 2

Let $I$ be an open interval, and $f,g$ differentiable on $I$. Then, for some constant $C$, $f^{\prime }(x)=g^{\prime }(x)$ if and only if $f(x)=g(x)+C$.

This will be really important in integration!

Corollary 3

Let $f:I\to \mathbb{R}$ be differentiable on $I$. For all $x\in I$, if $f^{\prime }(x)>0$ at $x\in I$, then $f$ is strictly increasing on $I$.

Cauchy MVT §

Below, we discuss a generalization of the Mean Value Theorem onto parametric curves.

Theorem: Cauchy Mean-Value Theorem

Suppose $f$ and $g$ are continuous and differentiable on $(a,b)$, creating a parametric curve $(f(t),g(t))$. If $g^{\prime }(t)\ne 0$, then $\forall t\in (a,b)$, $\exists t_0\in (a,b)$ such that

$$\frac{f^{\prime }(t_0)}{g^{\prime }(t_0)}=\frac{f(b)-f(a)}{g(b)-g(a)}$$

Note that this is a generalization, as we can set $g(t)=t$ for any suitable $f$ to prove the Mean Value Theorem on $f$.

Proof (Sketch)

Let $h(t)=f(t)-mg(t)$.

Choose $m$ such that $h(a)=h(b)$ in order to apply Rolle’s theorem to $h(t)$. We then follow a similar process to the MVT theorem proof.

This theorem is quite important! It sets the stage for the Taylor series Lagrange remainder theorem.

Theorem: Function Value Theorem

Let $I$ be an open interval, let $n\in \mathbb{N}$, and suppose we have a differentiable function $f:I\to \mathbb{R}$ which is $n$-times differentiable.

Suppose that at $x_0$

$$f(x_0)=f^{\prime }(x_0)=f^{\prime \prime }(x_0)=\cdots =f^{n-1}(x_0)=0$$

Then $\forall x\in I/\{x_0\}$, there exists a $z$ between $x_0$ and $x$ such that

$$f(x)=\frac{f^n(z)}{n!}(x-x_0{)}^n$$

Proof

Let $g(x)=(x-x_0{)}^n$. We have

$$\begin{array}{rl}& g(x_0)=0\\ & g^{\prime }(x_0)=n(x-x_0{)}^{n-1}{|}_{x=x_0}=0\\ & g^{\prime \prime }(x_0)=n(n-1)(x-x_0{)}^{n-2}{|}_{x=x_0}=0\\ & ⋮\\ & g^{n-1}(x_0)=n!(x-x_0){|}_{x=x_0}=0\\ & g^n(x_0)=n!\end{array}$$

Then, because $f(x_0)=g(x_0)=0$, we take (continuously applying to the $n^{th}$ derivative)

$$\begin{array}{rlr}\frac{f(x)}{g(x)}& =\frac{f(x)-f(x_0)}{g(x)-g(x_0)}=\frac{f^{\prime }(x_1)}{g^{\prime }(x_1)}& x_1\text{ between }{x}_0,x\\ & \frac{f^{\prime }(x_1)-f^{\prime }(x_0)}{g^{\prime }(x_1)-g^{\prime }(x_0)}=\frac{f^{\prime \prime }f(x_2)}{g^{\prime \prime }(x_2)}& \\ & ⋮& \\ & \frac{f^n(z)}{n!}& \end{array}$$

So, we find

$$f(x)=g(x)\frac{f^n(z)}{n!}=\frac{f^n(z)}{n!}(x-x_0{)}^n$$

Example: Remainder Theorem

Let $f:\mathbb{R}\to \mathbb{R}$ and $f(2)=f^{\prime }(2)=0$, and $|f^{\prime \prime }(x)|\le 3$ for all $x$. Give a bound on $f(5)$.

By the remainder theorem,

$$\begin{array}{r}f(5)=\frac{f^2(z)}{2!}(5-2{)}^2\\ |f(5)|\le \frac{3}{2}\cdot 9\le \frac{27}{2}\end{array}$$