MATH 409 Lecture 14

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Lecture Slides

Part 3: Differential and Integral Calculus

Derivative

A function $f$ is said to be differentiable at a point $a\in \mathbb {R}$ if it is defined on an open interval containing $a$ and the following limit exists

$\lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}$

The limit is denoted $f'(a)$ and called the derivative of $f$ at $a$.

An equivalent condition is

$f'(a)=\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}$

Examples

Constant function. $f(x)=c$ for $x\in \mathbb {R}$.

${\frac {f(x+h)-f(x)}{h}}={\frac {c-c}{h}}=0$ for all $x\in \mathbb {R}$ and $h\neq 0$.

Therefore the limit is 0 and $f$ is differentiable on $\mathbb {R}$ and $f'(x)=0$ for all $x\in \mathbb {R}$.

Identity function. $f(x)=x$ for $x\in \mathbb {R}$.

${\frac {f(x+h)-f(x)}{h}}={\frac {x+h-x}{h}}=1$ for all $x\in \mathbb {R}$, $h\neq 0$.

Therefore $f$ is differentiable on $\mathbb {R}$ and $f'(x)=1$ for all $x\in \mathbb {R}$.

Quadratic function. $f(x)=x^{2}$ for $x\in \mathbb {R}$.

${\frac {f(x+h)-f(x)}{h}}={\frac {(x+h)^{2}-x^{2}}{h}}={\frac {2xh+h^{2}}{h}}=2x+h$.

Therefore $\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}=\lim _{h\to 0}(2x+h)=2x$ (limit of continuous function)

Therefore $f$ is differentiable on $\mathbb {R}$ and $f'(x)=2x$ for $x\in \mathbb {R}$.

Harmonic function. $f(x)={\frac {1}{x}}$, $x\in (-\infty ,0)\cup (0,\infty )$.

${\frac {f(x+h)-f(x)}{h}}={\frac {1}{h}}\cdot \left({\frac {1}{x+h}}-{\frac {1}{x}}\right)={\frac {1}{h}}\cdot {\frac {x-(x+h)}{(x+h)\,x}}=-{\frac {1}{(x+h)\,x}}$.

Therefore $\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}=\lim _{h\to 0}-{\frac {1}{(x+h)\,x}}=-{\frac {1}{x^{2}}}$.

That is, $f$ is differentiable on $\mathbb {R} \setminus \left\{0\right\}$ and $f'(x)=-{\frac {1}{x^{2}}}$ for all $x\neq 0$.

Square root. $f(x)={\sqrt {x}}$, where $x\in [0,\infty )$.

${\frac {f(x+h)-f(x)}{h}}={\frac {{\sqrt {x+h}}-{\sqrt {x}}}{h}}\cdot {\frac {{\sqrt {x+h}}+{\sqrt {x}}}{{\sqrt {x+h}}+{\sqrt {x}}}}={\frac {(x+h)-(x)}{h({\sqrt {x+h}}+{\sqrt {x}})}}={\frac {1}{{\sqrt {x+h}}+{\sqrt {x}}}}$.

Therefore $\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}=\lim _{h\to 0}{\frac {1}{{\sqrt {x+h}}+{\sqrt {x}}}}={\frac {1}{2{\sqrt {x}}}}$.

In the case $x=0$,

$\lim _{h\to 0^{+}}{\frac {f(h)-f(0)}{h}}=\lim _{h\to 0}{\frac {1}{\sqrt {h}}}=+\infty$.

Hence $f$ is differentiable on $(0,\infty )$, and $f'(x)={\frac {1}{2{\sqrt {x}}}}$ for all $x>0$.

Sine function. $f(x)=\sin {x}$ for all $x\in \mathbb {R}$.

Using the formula $\sin {\alpha }-\sin {\beta }=2\,\sin {\frac {\alpha -\beta }{2}}\,\cos {\frac {\alpha +\beta }{2}}$, we obtain

${\frac {f(x+h)-f(x)}{h}}={\frac {\sin {(x+h)}-\sin {x}}{h}}={\frac {2}{h}}\,\sin {\frac {h}{2}}\,\cos {\frac {2x+h}{2}}$

Therefore

${\begin{aligned}\lim _{h\to 0}{\frac {2}{h}}\,\sin {\frac {h}{2}}\,\cos {\frac {2x+h}{2}}&=\lim _{h\to 0}{\frac {\sin {\frac {h}{2}}}{\frac {h}{2}}}\cdot \lim _{h\to 0}\cos {\left(x+{\frac {h}{2}}\right)}\\&=1\cdot \cos {x}\end{aligned}}$

That is, $f$ is differentiable on $\mathbb {R}$ and $f'(x)=\cos {x}$ for all $x\in \mathbb {R}$.

Differentiability Theorems

Differentiability and Continuity

Theorem. If a function $f$ is differentiable at a point $a$, then it is continuous at $a$.

Proof.

${\begin{aligned}\lim _{x\to a}f(a)&=\lim _{x\to a}\left(f(x)+{\frac {f(x)-f(a)}{x-a}}\,(x-a)\right)\\&=\lim _{x\to a}f(a)+\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}\cdot \lim _{x\to a}(x-a)\\&=f(a)+f'(a)\cdot 0=f(a)\end{aligned}}$

quod erat demonstrandum

A simple statement that will be used many times:

Note: Similarly, if $f$ has a right-hand derivative at $a$, then $\lim _{x\to a^{+}}f(x){=}f(a)$.
If $f$ has a left-hand derivative at $a$, then $\lim _{x\to a^{-}}f(x){=}f(a)$.

Sum Rule and Homogeneous Rule

Theorem. [Sum rule]. If functions $f$ and $g$ are differentiable at a point $a\in \mathbb {R}$, then the sum $f+g$ is also differentiable at $a$. Moreover, $(f+g)'(a)=f'(a)+g'(a)$.

Proof.

${\begin{aligned}\lim _{x\to a}{\frac {(f+g)(x)-(f+g)(a)}{x-a}}&=\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}+\lim _{x\to a}{\frac {g(x)-g(a)}{x-a}}\\&=f'(a)+g'(a)\end{aligned}}$

quod erat demonstrandum

It's worth noting that if the domains of $f$ and $g$ are not equal, then the domain of the sum of the functions is defined where the functions overlap. The sum over this open interval is well defined.

Theorem. [Homogeneous rule]. If a function $f$ is differentiable at a point $a\in \mathbb {R}$, then for any $r\in \mathbb {R}$ the scalar multiple $r\,f$ is also differentiable at $a$. Moreover, $(r\,f)'(a)=r\,f'(a)$.

Proof.

${\begin{aligned}\lim _{x\to a}{\frac {(r\,f)(x)-(r\,f)(a)}{x-a}}&=\lim _{x\to a}r\,{\frac {f(x)-f(a)}{x-a}}\\&=r\,f'(a)\end{aligned}}$

quod erat demonstrandum

Product Rule

Theorem. [Product rule]. If the functions $f$ and $g$ are differentiable at a point $a\in \mathbb {R}$, then the product $f\cdot g$ is also differentiable at $a$. Moreover, $(f\cdot g)'(a)=f'(a)\,g(a)+f(a)\,g'(a)$.

Proof. Since $f$ and $g$ are differentiable at $a$, there is an open interval $I=(c,d)$ containing $a$ such that both $f$ and $g$ are defined on $I$. (see note above regarding sums) For every $x\in I\setminus \left\{a\right\}$ we have:

${\begin{aligned}f(x)\,g(x)-f(a)\,g(a)&=f(x)\,g(x)-f(a)\,g(a)+f(a)\,g(x)-f(a)\,g(a)\\&=(f(x)-f(a))\,g(x)+f(a)\,(g(x)-g(a))\end{aligned}}$

Then ${\frac {(f\cdot g)(x)-(f\cdot g)(a)}{x-a}}={\frac {f(x)-f(a)}{x-a}}\,g(x)+f(a)\,{\frac {g(x)-g(a)}{x-a}}$ so that

${\begin{aligned}\lim _{x\to a}{\frac {(f\cdot g)(x)-(f\cdot g)(a)}{x-a}}&=\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}\cdot \lim _{x\to a}g(x)+\lim _{x\to a}f(a)\cdot \lim _{x\to a}{\frac {g(x)-g(a)}{x-a}}\\&=f'(x)\,g(a)+f(a)\,g'(a)\end{aligned}}$

We used the fact that $f$ and $g$ are continuous at $a$ to evaluate the limits in the last step.

quod erat demonstrandum

Reciprocal Rule

Theorem. [Reciprocal rule]. If a function $f$ is differentiable at a point $a\in \mathbb {R}$ and $f(a)\neq 0$, then the function ${\frac {1}{f}}$ is also differentiable at $a$. Moreover, $\left({\frac {1}{f}}\right)'(a)=-{\frac {f'(a)}{f^{2}(a)}}$.

Proof. The function $f$ is defined on an open interval $(c,d)$ containing $a$. We know that $f$ is continuous at $a$. Since $\epsilon =\left|f(a)\right|>0$, there exists $\delta >0$ such that $\left|f(x)-f(a)\right|<\epsilon$ for any $x\in I=(c,d)\cap (a-\delta ,a+\delta )$. Then $f(x)\neq 0$ for all $x\in I$. In particular, ${\frac {1}{f}}$ is defined on $I$, an open interval containing $a$.

Now

${\begin{aligned}\lim _{x\to a}{\frac {\left({\frac {1}{f}}\right)(x)-\left({\frac {1}{f}}\right)(a)}{x-a}}&=\lim _{x\to a}\left({\frac {1}{f(x)}}-{\frac {1}{f(a)}}\right)\,{\frac {1}{x-a}}\\&=\lim _{x\to a}{\frac {f(a)-f(x)}{f(x)\,f(a)}}\,{\frac {1}{x-a}}\\&=\lim _{x\to a}\left(-{\frac {f(x)-f(a)}{x-a}}\,{\frac {1}{f(x)\,f(a)}}\right)\\&=-\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}\,\lim _{x\to a}{\frac {1}{f(x)\,f(a)}}\\&={\frac {f'(a)}{f^{2}(a)}}\end{aligned}}$

quod erat demonstrandum

Difference Rule and Quotient Rule

Theorem. [Difference rule]. If $f$ and $g$ are differentiable at a point $a\in \mathbb {R}$, then the difference $(f-g)$ is also differentiable at $a$. Moreover, $(f-g)'(a)=f'(a)-g'(a)$.

Proof. By the Homogeneous Rule, the function $-g=(-1)\,g$ is differentiable at $a$ and $(-g)'(a)=-g'(a)$ By the Sum Rule, $f-g=f+(-g)$ is also differentiable at $a$ and $(f-g)'(a)=f'(a)+(-g)'(a)=f'(a)-g'(a)$.

quod erat demonstrandum

Theorem. [Quotient rule]. If functions $f$ and $g$ are differentiable at $a\in \mathbb {R}$ and $g(a)\neq 0$, then the quotient ${\frac {f}{g}}$ is also differentiable at $a$. Moreover, $\left({\frac {f}{g}}\right)'(a)={\frac {f'(a)\,g(a)-f(a)\,g'(a)}{g^{2}(a)}}$.

Proof. By the Reciprocal Rule, the function ${\frac {1}{g}}$ is differentiable at $a$ and $\left({\frac {1}{g}}\right)'(a)={\frac {-g'(a)}{g^{2}(a)}}$. By the product rule, the function ${\frac {f}{g}}=f\cdot {\frac {1}{g}}$ is also differentiable at $a$ and

${\begin{aligned}\left({\frac {f}{g}}\right)'(a)&={\frac {f'(a)}{g(a)}}+f(a)\,\left({\frac {1}{g}}\right)'(a)\\&={\frac {f'(a)\,g(a)-f(a)\,g'(a)}{g^{2}(a)}}\end{aligned}}$

quod erat demonstrandum

Chain Rule

Theorem. [Chain rule]. If a function $f$ is differentiable at a point $a\in \mathbb {R}$ and a function $g$ is differentiable at $f(a)$, then the composition $g\circ f$ is differentiable at $a$. Moreover, $(g\circ f)'(a)=g'(f(a))\cdot f'(a)$.

Proof. The function $f$ is defined on an open interal $I=(a-\delta ,a+\delta )$ while $g$ is defined on an open interal $J=(f(a)-\epsilon ,f(a)+\epsilon )$. Since $f$ is continuous at $a$ (because it is differentiable), there exists $\delta _{0}\in (0,\delta )$ such that $f(I_{0})\subset J$, where $I_{0}=(a-\delta _{0},a+\delta _{0})$. Then $g\circ f$ is defined on $I_{0}$. For any $x\in I_{0}$ such that $f(x)\neq f(a)$,

${\frac {(g\circ f)(x)-(g\circ f)(a)}{x-a}}={\frac {g(f(x))-g(f(a))}{f(x)-f(a)}}\cdot {\frac {f(x)-f(a)}{x-a}}$

This implies the Chain Rule unless there is a sequence $\left\{x_{n}\right\}$ converging to $a$ such that $x_{n}\neq a$ while $f(x_{n})=f(a)$. In this case, $(g\circ f)'(a)=f'(a)=0$.

quod erat demonstrandum

Examples

Cosine function. $f(x)=\cos {x}$ for $x\in \mathbb {R}$.

The function $f$ can be represented as a composition $f=h\circ g$, where $g(x)=x+{\frac {\pi }{2}}$ and $h(x)=\sin {x}$ for all $x\in \mathbb {R}$. Since $g'(x)=1$ and $h'(x)=\cos {x}$ for all $x\in \mathbb {R}$, the Chain rule implies that $f$ is differentiable on $\mathbb {R}$ and $f'(x)=h'(g(x))\,g(x)=\cos {\left(x+{\frac {\pi }{2}}\right)}=-\sin {x}$ for all $x\in \mathbb {R}$.

Tangent function. $f(x)=\tan {x}$ for $x\in \left(-{\frac {\pi }{2}},{\frac {\pi }{2}}\right)$.

Since $f(x)={\frac {\sin {x}}{\cos {x}}}$ and $\cos {x}\neq 0$ for all $x\in \left(-{\frac {\pi }{2}},{\frac {\pi }{2}}\right)$, the Quotient Rule implies that $f$ is differentiable on $\left(-{\frac {\pi }{2}},{\frac {\pi }{2}}\right)$ and

$f'(x)={\frac {(\sin {x})'\,\cos {x}-\sin {x}\,(\cos {x})'}{\cos ^{2}{x}}}={\frac {\cos ^{2}{x}+\sin ^{2}{x}}{\cos ^{2}{x}}}={\frac {1}{\cos ^{2}{x}}}$

For all $x\in \left(-{\frac {\pi }{2}},{\frac {\pi }{2}}\right)$.