MA5131 Summary 2

Definition of Derivative

The derivative is defined as $f^{\prime }(a)=\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}=\underset{x\to 0}{lim}\frac{f(a+x)-f(a)}{x}$.

Intuitively, the derivative is measuring the slope of a line at a instantaneous point. So we can define the derivative at $x=a$ as taking the gradient of the $2$ points $(a,f(a))$ and $(b,f(b))$ as $b$ gets closer and closer to $a$, which is what the above formula describes.

If the limit $\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}$ exists, then $f^{\prime }(a)$ is well-defined and we say that $f$ is differentiable at $a$. By this definition, it is a consequence that if a function is only defined on $[l,r]$ it is not differentiable on $l$ and $r$ as the limit does not exist.

Using the above definitions, we can find the derivative of functions. Doing so is known as Differentiation from First Principles.

Here are some examples:

$\frac{d}{dx}x=\underset{a\to 0}{lim}\frac{(x+a)-x}{a}=\underset{a\to 0}{lim}\frac{a}{a}=1$

$\frac{d}{dx}{x}^n=\underset{a\to 0}{lim}\frac{(x+a{)}^n-x^n}{a}=\underset{a\to 0}{lim}\frac{(\genfrac{}{}{0ex}{}{n}{1})x^{n-1}a+O(a^2)}{a}=(\genfrac{}{}{0ex}{}{n}{1})x^{n-1}=n{x}^{n-1}$, where $n$ is a positive integer

$$\begin{array}{rl}\frac{d}{dx}\sin (x)=& \underset{a\to 0}{lim}\frac{sin(x+a)-sin(x)}{a}=\underset{a\to 0}{lim}\frac{sin((x+\frac{a}{2})+\frac{a}{2})-sin((x+\frac{a}{2})-\frac{a}{2})}{a}\\ \\ =& \underset{a\to 0}{lim}\frac{2\cos (x+\frac{a}{2})\sin (\frac{a}{2})}{a}=\underset{a\to 0}{lim}\cos (x+\frac{a}{2})\cdot \underset{a\to 0}{lim}\frac{\sin (\frac{a}{2})}{\frac{a}{2}}=\underset{a\to 0}{lim}\cos (x+\frac{a}{2})=\cos (x)\end{array}$$

Estimating Derivative given Table of Values

When we need to find a derivative a point given a few datapoints, it is recommended to use $f^{\prime }(a)\approx \frac{f(a+h)-f(a-h)}{2h}$ unless $a$ is a endpoint, then we have no choice but to use $f^{\prime }(a)\approx \frac{f(a+h)}{h}$.

If $f$ is differentiable at $a$, then $f$ is continuous at $a$.

Proof: Since $f^{\prime }(a)$ is well-defined, $\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}$ exists.

We wish to show that $\underset{x\to a}{lim}f(x)=f(a)$.

$$\begin{array}{rl}\underset{x\to a}{lim}f(x)& =\underset{x\to a}{lim}(f(x)-f(a)+f(a)),\text{since }f(a)\text{ is well-defined}\\ \\ & =\underset{x\to a}{lim}(\frac{f(x)-f(a)}{x-a}\cdot (x-a))+\underset{x\to a}{lim}f(a)\\ \\ & =f^{\prime }(0)\cdot 0+f(a)\\ \\ & =f(a)\end{array}$$

Common Derivatives

$\frac{d}{dx}c=0$, where $c$ is a constant	$\frac{d}{dx}\sin (x)=\cos (x)$
$\frac{d}{dx}{x}^n=n{x}^{n-1}$, $n\ne 0$	$\frac{d}{dx}\cos (x)=-\sin (x)$
$\frac{d}{dx}\ln x=\frac{1}{x}$	$\frac{d}{dx}\arcsin (x)=\frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx}{e}^x=e^x$	$\frac{d}{dx}\arctan (x)=\frac{1}{1+x^2}$

Rules

$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$ (Chain Rule)

$\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$ (Product rule)

Derivative of Inverse Function

$(f^{-1}{)}^{\prime }(a)=\frac{1}{f^{\prime }(f^{-1}(a))},f^{\prime }(f^{-1}(a))\ne 0$

The derivative of $f^{-1}$ at $(a,f^{-1}(a))$, it is related to the derivative of $f$ at $(f^{-1}(a),a)$ by a reciprocal relationship.

Rolle’s Theorem

If $f$ is a function that is both continuous on $[a,b]$, differentiable on the interval $(a,b)$ and $f(a)=f(b)$, then there exists $c\in (a,b)$ such that $f^{\prime }(c)=0$.

Note that the differentiability condition is important here. Consider the function $f(x)=|x|$ on the interval $[-1,1]$. It can be shown that it is continuous on $[-1,1]$ and $f(-1)=f(1)$. However, there is no point where its derivative is $0$.

Mean Value Theorem

This is a corollary of Rolle’s Theorem.

If $f$ is a function that is both continuous on $[a,b]$ and differentiable on the interval $(a,b)$, then there exists $c\in (a,b)$ such that $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$.

The intuitive proof of this using Rolle’s theorem is that we shift the a function such that $f(a)=f(b)$ by a linear function.

Increasing and Decreasing Functions

$f$ is non-decreasing on $I$ if $f(a)\le f(b)$ for $a,b\in I$ where $a<b$.
$f$ is increasing on $I$ if $f(a)<f(b)$ for $a,b\in I$ where $a<b$.
$f$ is non-increasing on $I$ if $f(a)\ge f(b)$ for $a,b\in I$ where $a<b$.
$f$ is decreasing on $I$ if $f(a)>f(b)$ for $a,b\in I$ where $a<b$.
$f$ is constant on $I$ if $f(a)=f(b)$ for $a,b\in I$ where $a<b$.

Interestingly, a single point is defined to be both increasing and decreasing by this definition.

Corollaries of Mean Value Theorem

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$:
(a) If $f^{\prime }(x)\ge 0$ for all $x\in (a,b)$, then $f$ is non-decreasing on $[a,b]$.
(b) If $f^{\prime }(x)>0$ for all $x\in (a,b)$, then $f$ is increasing on $[a,b]$.
(c) If $f^{\prime }(x)\le 0$ for all $x\in (a,b)$, then $f$ is non-increasing on $[a,b]$.
(d) If $f^{\prime }(x)<0$ for all $x\in (a,b)$, then $f$ is decreasing on $[a,b]$.
(e) If $f^{\prime }(x)=0$ for all $x\in (a,b)$, then $f$ is constant on $[a,b]$.

The converse of (a),(c) and (e) are true, but the converse of (b) and (d) is false.

Counter example of converse of (b)

$x^3$ is increasing on $[-1,1]$, but $\frac{d}{dx}{x}^3{|}_{x=0}=0$.

Proof of (a)

Suppose that $f$ is decreasing on $[a,b]$, that is there exists $c,d\in [a,b]$ such that $c<d$ and $f(c)>f(d)$.

By Mean Value Theorem, there exists $x\in (c,d)$ such that $f^{\prime }(x)=\frac{f(d)-f(c)}{d-c}<0$. However, $f^{\prime }(x)\ge 0$ for all $x\in (a,b)$.

This is a contradiction. Therefore, (a) is true.

Proof of converse of (a) (not rigorous)

Suppose there exists $x\in (a,b)$ such that $f^{\prime }(x)<0$. This implies that $\underset{h\to x^{+}}{lim}\frac{f(h)-f(x)}{h-x}<0$.

Since $h-x>0$, this means that for a small interval $h\in (x,x+ϵ]$, $f(h)-f(x)<0$ or that $f(x)>f(h)$. However, we have $f(x)\le f(h)$ for $x,h\in [a,b]$ and $x<h$.

This is a contradiction. Therefore, converse of (a) is true.

Critical Points

Let $c$ be an interior point (not endpoint if domain if closed). If $f^{\prime }(c)=0$ or $f^{\prime }(c)$ does not exist, then $(c,f(c))$ is a critical point.

First Derivative Test

If $f$ is differentiable on a small open interval around $c$ with exclusion of $c$
(i) If $f^{\prime }$ changes sign from negative to positive, local minimum
(ii) If $f^{\prime }$ changes sign from positive to negative, local maximum
(iii) Else if $f^{\prime }\ne 0$, it is a inflection point

Second Derivative Test

Suppose $f^{\prime \prime }(c)=0$
(i) If $f^{\prime \prime }(c)>0$, local minimum
(ii) If $f^{\prime \prime }(c)<0$, local maximum
(iii) Else, inconclusive

Concavity

If $f$ is differentiable on a open interval $I$.
(i) If $f$ is increasing, concave upwards
(ii) If $f$ is decreasing, concave downwards

Suppose $f$ is twice differentiable on a open interval $I$.
(i) If $f^{\prime \prime }>0$, concave upwards
(ii) If $f^{\prime \prime }<0$, concave downwards

However, the converse is not true. $x^3$ is concave upwards on $(-1,1)$ but $\frac{d^2}{d{x}^2}{x}^3{|}_{x=0}=0$.

Inflection Point

$(c,f(c))$ is a inflection point if $f$ is continuous at $c$ and the concavity changes around $c$.

Note that $f^{\prime \prime }(c)=0$ does not imply inflection point. $x^3$ is concave upwards on $(-1,1)$ but $\frac{d^2}{d{x}^2}{x}^3{|}_{x=0}=0$.

L’Hopital’s Rule

$\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$ if the original limit is of the form $\frac{0}{0}$ or $\pm \frac{\mathrm{\infty }}{\mathrm{\infty }}$.

Other indeterminant forms such as $0\cdot \mathrm{\infty }$, $\mathrm{\infty }-\mathrm{\infty }$, $0^0$, ${\mathrm{\infty }}^{\mathrm{\infty }}$,$1^{\mathrm{\infty }}$ can be turned in $\frac{0}{0}$ or $\pm \frac{\mathrm{\infty }}{\mathrm{\infty }}$ using algebraic manipulation to apply L’Hopital’s Rule.

Growth of Functions

Order from slowest to fastest growth: constant, logarithmic, polynomial, exponential, factorial.

---

<

Previous Post

MA5131 Summary 1

>

Next Post

MA5131 Summary 3