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

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

The limit is denoted $f^{\prime }(a)$ and called the derivative of $f$ at $a$.

An equivalent condition is

$f^{\prime }(a)=\underset{x\to a}{\lim }\frac{f(x)-f(a)}{x-a}$

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 $h\ne 0$.

Therefore the limit is 0 and $f$ is differentiable on $\mathbb{ℝ}$ and $f^{\prime }(x)=0$ 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 $x\in \mathbb{ℝ}$, $h\ne 0$.

Therefore $f$ is differentiable on $\mathbb{ℝ}$ and $f^{\prime }(x)=1$ 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 $\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}=\underset{h\to 0}{\lim }(2x+h)=2x$ (limit of continuous function)

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

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

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

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

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

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

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

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

In the case $x=0$,

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

Hence $f$ is differentiable on $(0,\mathrm{\infty })$, and $f^{\prime }(x)=\frac{1}{2\sqrt{x}}$ 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

$\begin{array}{rl}\underset{h\to 0}{\lim }\frac{2}{h}\sin \frac{h}{2}\cos \frac{2x+h}{2}& =\underset{h\to 0}{\lim }\frac{\sin \frac{h}{2}}{\frac{h}{2}}\cdot \underset{h\to 0}{\lim }\cos (x+\frac{h}{2})\\ & =1\cdot \cos x\end{array}$

That is, $f$ is differentiable on $\mathbb{ℝ}$ and $f^{\prime }(x)=\cos x$ 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.

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

quod erat demonstrandum

A simple statement that will be used many times:

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

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, $(f+g{)}^{\prime }(a)=f^{\prime }(a)+g^{\prime }(a)$.

Proof.

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

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 $rf$ is also differentiable at $a$. Moreover, $(rf{)}^{\prime }(a)=r{f}^{\prime }(a)$.

Proof.

$\begin{array}{rl}\underset{x\to a}{\lim }\frac{(rf)(x)-(rf)(a)}{x-a}& =\underset{x\to a}{\lim }r\frac{f(x)-f(a)}{x-a}\\ & =r{f}^{\prime }(a)\end{array}$

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, $(f\cdot g{)}^{\prime }(a)=f^{\prime }(a)g(a)+f(a)g^{\prime }(a)$.

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:

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

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

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

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 $f(a)\ne 0$, then the function $\frac{1}{f}$ is also differentiable at $a$. Moreover, $\left(\frac{1}{f}\right)(a)=-\frac{f^{\prime }(a)}{f^2(a)}$.

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

Now

$\begin{array}{rl}\underset{x\to a}{\lim }\frac{\left(\frac{1}{f}\right)(x)-\left(\frac{1}{f}\right)(a)}{x-a}& =\underset{x\to a}{\lim }(\frac{1}{f(x)}-\frac{1}{f(a)})\frac{1}{x-a}\\ & =\underset{x\to a}{\lim }\frac{f(a)-f(x)}{f(x)f(a)}\frac{1}{x-a}\\ & =\underset{x\to a}{\lim }(-\frac{f(x)-f(a)}{x-a}\frac{1}{f(x)f(a)})\\ & =-\underset{x\to a}{\lim }\frac{f(x)-f(a)}{x-a}\underset{x\to a}{\lim }\frac{1}{f(x)f(a)}\\ & =\frac{f^{\prime }(a)}{f^2(a)}\end{array}$

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, $(f-g{)}^{\prime }(a)=f^{\prime }(a)-g^{\prime }(a)$.

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

quod erat demonstrandum

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

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

$\begin{array}{rl}\left(\frac{f}{g}\right)(a)& =\frac{f^{\prime }(a)}{g(a)}+f(a)\left(\frac{1}{g}\right)(a)\\ & =\frac{f^{\prime }(a)g(a)-f(a)g^{\prime }(a)}{g^2(a)}\end{array}$

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, $(g\circ f{)}^{\prime }(a)=g^{\prime }(f(a))\cdot f^{\prime }(a)$.

Proof. The function $f$ is defined on an open interal $I=(a-\delta ,a+\delta )$ while $g$ is defined on an open interal $J=(f(a)-ϵ,f(a)+ϵ)$. 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 $f(x)\ne f(a)$,

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

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

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 $g^{\prime }(x)=1$ and $h^{\prime }(x)=\cos x$ for all $x\in \mathbb{ℝ}$, the Chain rule implies that $f$ is differentiable on $\mathbb{ℝ}$ and $f^{\prime }(x)=h^{\prime }(g(x))g(x)=\cos (x+\frac{\pi }{2})=-\sin x$ for all $x\in \mathbb{ℝ}$.

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

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

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

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