Differential Calculus : Differentiation: First Principles

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Differentiation: First Principles

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Overview

Differentiation : First Principles

» cause-effect relation in two quantities

    → eg: speed-displacement

» The effect is calculated as a function of an algebraic expression in a variable.

    → eg: displacement $s = 3 t^{2} + 2$ $s=3t^2+2$

» The cause is derived to be "rate of change of effect with respect to the variable".

    → eg : Speed = rate of change of displacement

» In such a case, the cause is another algebraic expression in the variable.

    → speed is a function of $t$ $t$

» The cause is computed as rate of change : the change in effect for a small change $δ \to 0$ $delta->0$ in the variable.

    → speed $= lim_{δ \to 0} \frac{s (t + δ) - s (t)}{δ}$ $=lim_(delta->0) (s(t+delta)-s(t))/delta$

» Derivative or Differentiation of a function

For a small change in variable $x$ $x$ , the rate of change in the function $f (x)$ $f(x)$ is the derivative of the function.

$\frac{d}{d x} f (x)$ $d/(dx) f(x)$

$= f' (x)$ $= f′(x)$

$= lim_{δ \to 0} \frac{f (x + δ) - f (x)}{δ}$ $= lim_(delta->0) (f(x+delta) - f(x))/delta$

rate of change

A car travels a distance $20$ $20$ m in $2$ $2$ seconds. The speed of the car is " $20 m / 2 \sec = 10$ $20m // 2sec = 10$ m/sec"

Speed is computed as distance traveled (change) per unit time (rate).

Speed is rate of change of distance.

quantity as an expression

A car is moving at speed $20$ $20$ m/sec. What is the distance covered in $t$ $t$ sec ?

without a numerical value for $t$ $t$ the distance traveled can be given as an expression.

For any value of $t$ $t$ , the distance traveled is $s = 20 t$ $s=20t$ meter

Measurement can be expressed as a function of a variable.

rate of change of expression

A car is moving with displacement given as a function of time $s = 3 t^{2} + 2$ $s=3t^2+2$ . The distance at $t = 2 \sec$ $t=2sec$ is computed by substituting $t = 2$ $t=2$ in $s = 3 t^{2} + 2$ $s=3t^2+2$ $= 3 \times 2^{2} + 2 = 14 m$ $=3xx2^2+2 = 14m$

A car is moving with displacement given as a function of time $s = 3 t^{2} + 2$ $s=3t^2+2$ . Couple of students want to calculate the speed.
We know that $speed = \frac{distance}{time}$ $text(speed) = text(distance) / text(time)$

• Person A does the following: At $t = 2$ $t=2$ s, the distance traveled is $14$ $14$ m. So the speed is $\frac{14}{2} = 7$ $14/2 = 7$ m/sec.

• Person B does the following: At $t = 1$ $t=1$ s, the distance traveled is $5$ $5$ m. So the speed is $\frac{5}{1} = 5$ $5/1 = 5$ m/sec.

Which one is correct? "Neither of them". The answer is explained in the following.

Person A calculated average speed till $2$ $2$ sec

Person B calculated average speed till $1$ $1$ sec"

The results from these two are different and we understand that the average speed changes with time and we expect that "speed is a function of time".

Speed is rate of change of displacement. So, one has to figure out how to find the rate of change of an algebraic expression.

instantaneous

Consider the problem, a car is moving with displacement given as a function of time $s = 3 t^{2} + 2$ $s=3t^2+2$ . Speed is the rate of change. Few students are set to find rate of change

• Person A found rate of change for 1 second interval. $3 {(t + 1)}^{2} + 2 - 3 t^{2} - 2$ $3(t+1)^2+ 2- 3t^2-2$ $= 6 t + 3$ $=6t+3$

• Person B found rate of change for 2 second interval. $\frac{3 {(t + 2)}^{2} + 2 - 3 t^{2} - 2}{2}$ $(3(t+2)^2+ 2- 3t^2-2)/2$ $= 6 t + 6$ $=6t+6$

• Person C found rate of change for 0.5 second interval $\frac{3 {(t + .5)}^{2} - 3 t^{2}}{.5}$ $(3(t+.5)^2-3t^2)/(.5)$ $= 6 t + 1.5$ $=6t+1.5$

The above actually provides "Average of two values representing speed at $t$ $t$ and $t + δ$ $t+delta$ ". That is, for Person A, the $δ = 1$ $delta = 1$ , for person B, the $δ = 2$ $delta=2$ , and for person C, the $δ = 0.5$ $delta=0.5$ .

The calculation does not provide instantaneous speed. What is "instantaneous".

A speedometer is attached to a wheel of the car. The speedometer measures the speed at which the wheel rotates and proportionally provides the speed of the car. The speed shown in the speedometer is the "instantaneous speed for any given time".

when interval is 0

The average of speed between $t$ $t$ and $t + δ$ $t + delta$ is $\frac{s (t + δ) - s (t)}{δ}$ $(s(t+delta)-s(t))/delta$ $= 6 t + 3 δ$ $=6t+3delta$ . If the value of $δ$ $delta$ is small, then the accuracy of the instantaneous speed is better.

Instantaneous speed is when the time interval $δ$ $delta$ is zero. This is a big step in understanding. The displacement is continuously changing with time and at an instance the rate of change is calculated.

substitute 0

The instantaneous speed is $\frac{s (t + δ) - s (t)}{δ} ∣_{δ = 0}$ $(s(t+delta)-s(t))/delta |_(delta=0)$ . The value of this calculation when $δ = 0$ $delta=0$ is substituted is "indeterminate value $0 / 0$ $0//0$ ".

To solve a function evaluating to indeterminate value "Use Limit of the function as $δ$ $delta$ approaching $0$ $0$ ."

limit

A car is moving with displacement given as a function of time $s (t) = 3 t^{2} + 2$ $s(t)=3t^2+2$ . The instantaneous speed is
$\frac{s (t + δ) - s (t)}{δ} ∣_{δ = 0}$ $(color(deepskyblue)(s(t+delta))-color(coral)(s(t)))/delta |_(delta=0)$
$= \frac{s (t) - s (t)}{δ}$ $= (s(t)-s(t))/delta$
$= \frac{0}{0}$ $=0/0$

Since the speed evaluates to indeterminate value, the limit is used:

$lim_{δ \to 0} \frac{s (t + δ) - s (t)}{δ}$ $lim_(delta->0) (color(deepskyblue)(s(t+delta))-color(coral)(s(t)))/delta$

$= lim_{δ \to 0} \frac{3 {(t + δ)}^{2} + 2 - 3 t^{2} - 2}{δ}$ $quad quad = lim_(delta->0)(color(deepskyblue)(3(t+delta)^2 + 2) - color(coral)(3t^2 -2) )/delta$

canceling $2$ $color(deepskyblue)(2)$ and $- 2$ $color(coral)(-2)$ and expanding the square
$= lim_{δ \to 0} \frac{3 t^{2} + 6 t δ + 3 δ^{2} - 3 t^{2}}{δ}$ $quad quad = lim_(delta->0) (color(deepskyblue)(3t^2+6t delta + 3delta^2) -color(coral)(3t^2))/delta$

canceling $3 t^{2}$ $color(deepskyblue)(3t^2)$ and $- 3 t^{2}$ $color(coral)(-3t^2)$
$= lim_{δ \to 0} \frac{6 t δ + 3 δ^{2}}{δ}$ $quad quad = lim_(delta->0) color(deepskyblue)(6t delta + 3delta^2)/delta$

canceling $δ$ $delta$ from numerator and denominator
$= lim_{δ \to 0} 6 t + 3 δ$ $quad quad =lim_(delta->0) color(deepskyblue)(6t + 3delta)$

applying limit.
$= 6 t$ $quad quad = color(deepskyblue)(6t)$

The instantaneous speed is computed as an algebraic expression.

summary

Summarizing the learning so far:

• Two quantities are in a cause-effect relation.

• The effect is calculated as a function of an algebraic expression in a variable.

• The cause is derived to be "rate of change of effect with respect to the variable".
(note: there are other forms of relation between cause-effect, such as multiple, addition, exponent. In this topic, we are concerned with only the rate of change relation.)

• In such a case, the cause is another algebraic expression in the variable.

• The cause is computed as rate of change : the change in effect for a small change $δ \to 0$ $delta->0$ in the variable.

This calculation is named as differentiation or derivative of the function.

when to?

Differentiation in the context of cause-effect relation: If effect is given by $f (x)$ $f(x)$ and the cause is the rate-of-change, then the cause is computed as differentiation or derivative of $f (x)$ $f(x)$ denoted as $\frac{d}{d x} f (x)$ $d/(dx) f(x)$ or $f' (x)$ $f′(x)$ .

$\frac{d}{d x} f (x) = lim_{δ \to 0} \frac{f (x + δ) - f (x)}{δ}$ $d/(dx) f(x) = lim_(delta->0) (f(x+delta)-f(x))/delta$ . It is duly noted that the cause-effect can be of various relationships, like sum, difference, product, etc. Only when the cause-effect is established to be rate-of-change, the differentiation is used to get instantaneous-rate-of-change.

Cause-effect is explained to understand the physical significance. Abstracting this and understanding the quantities involved in differentiation: A quantity $u = f (x)$ $u=f(x)$ is related to another quantity $v$ $v$ such that $v$ $v$ is the instantaneous rate of change of $u$ $u$ with respect to $x$ $x$ , then
$v = \frac{d u}{d x} = lim_{δ \to 0} \frac{f (x + δ) - f (x)}{δ}$ $v = (du)/(dx) = lim_(delta->0) (f(x+delta)-f(x))/delta$

Note that $\frac{d u}{d x}$ $(du)/(dx)$ is another quantity related to the given quantity $u$ $u$ .

various forms

The derivative of the function $y = f (x)$ $y=f(x)$ can be given in different forms:

For a small change in variable $x$ $x$ , the instantaneous rate of change of the function $y = f (x)$ $y=f(x)$ is the derivative of the function.

$\frac{d y}{d x}$ $(dy)/(dx)$

$= y'$ $=y′$

$= lim_{Δ x \to 0} \frac{Δ y}{Δ x}$ $= lim_(Delta x ->0) (Delta y)/(Delta x)$

$= \frac{d}{d x} f (x)$ $=d/(dx) f(x)$

$= f' (x)$ $= f′(x)$

$= lim_{δ \to 0} \frac{f (x + δ) - f (x)}{δ}$ $= lim_(delta->0) (f(x+delta) - f(x))/delta$

$= lim_{Δ x \to 0} \frac{f (x + Δ x) - f (x)}{Δ x}$ $= lim_(Delta x ->0) (f(x+Delta x) - f(x))/(Delta x)$

The instantaneous rate of change of a function is called "differentiation or derivative of the function".

Students can connect the notation $\frac{d}{d x} f (x)$ $d/(dx) f(x)$ as

• the small difference in $x$ $x$ is given by denominator $d x$ $dx$

• the effective difference (because of small difference in $x$ $x$ ) in the function is given by numerator $d$ $d$

• $\frac{d}{d x}$ $d/(dx)$ denotes the difference in function with respect to a small difference in $x$ $x$ .

The instantaneous rate of change of a function with respect to the variable is the derivative of the function. The word "instantaneous" may be dropped and it is often referred to as "rate of change".

examples

Finding the derivative of $y = x^{2} + x$ $y=x^2+x$ in first principles:

$\frac{d y}{d x}$ $(dy)/(dx)$

$= lim_{δ \to 0} [f (x + δ)$ $=lim_(delta->0) [color(coral)(f(x+delta))$ $- f (x)] / δ$ $- color(deepskyblue)(f(x))]//delta$

$= lim_{δ \to 0} [{(x + δ)}^{2} + x + δ$ $=lim_(delta->0) [color(coral)((x+delta)^2+ x+ delta)$ $- (x^{2} + x)] / δ$ $- color(deepskyblue)((x^2+x))]//delta$

$= lim_{δ \to 0} [x^{2} + 2 δ x + δ^{2} + x + δ$ $=lim_(delta->0) [color(coral)(x^2+2delta x + delta^2 + x + delta)$ $- x^{2} - x] / δ$ $- color(deepskyblue)(x^2 -x)]//delta$

$= lim_{δ \to 0} [2 δ x + δ^{2} + δ] / δ$ $=lim_(delta->0) [color(coral)(2delta x + delta^2 + delta)]//delta$

$= lim_{δ \to 0} 2 x + δ + 1$ $=lim_(delta->0) 2x+delta+1$

Substituting $δ = 0$ $delta=0$ ,
$\frac{d}{d x} (x^{2} + x) = 2 x + 1$ $d/(dx) (x^2+x) = 2x+1$

Finding the derivative of $y = \sin x$ $y=sin x$ in first principles:

$\frac{d y}{d x}$ $(dy)/(dx)$

$= lim_{δ \to 0} [f (x + δ)$ $=lim_(delta->0) [color(coral)(f(x+delta))$ $- f (x)] / δ$ $- color(deepskyblue)(f(x))]//delta$

$= lim_{δ \to 0} [\sin (x + δ)$ $=lim_(delta->0) [color(coral)(sin(x+delta)$ $- \sin x] / δ$ $- color(deepskyblue)(sinx)]//delta$

$= lim_{δ \to 0} [\sin x \cos δ + \cos x \sin δ$ $=lim_(delta->0) [color(coral)(sinx cos delta + cos x sin delta)$ $- \sin x] / δ$ $-color(deepskyblue)(sinx)]//delta$

$= lim_{δ \to 0} \frac{\cos x \sin δ}{δ}$ $= lim_(delta->0) [color(coral)(cos x sin delta)]/delta$ $- lim_{δ \to 0} \frac{\sin x - \sin x \cos δ}{δ}$ $- lim_(delta->0) (color(deepskyblue)(sin x) - color(coral)(sin x cos delta))/delta$

$= \cos x lim_{δ \to 0} \frac{\sin δ}{δ}$ $= cos x lim_(delta->0) (sin delta)/delta$ $- \sin x lim_{δ \to 0} \frac{1 - \cos δ}{δ}$ $- sin x lim_(delta->0)(1 - cos delta)/delta$

applying the standard limits
$= \cos x \times 1 - \sin x \times 0$ $= cos x xx 1 - sin x xx 0$

That is
$\frac{d}{d x} \sin x = \cos x$ $d/(dx) sin x = cos x$ "

summary

Derivative or Differentiation of a function : For a small change in variable $x$ $x$ , the instantaneous rate of change in the function $f (x)$ $f(x)$ is the derivative of the function.

$\frac{d}{d x} f (x)$ $d/(dx) f(x)$

$= f' (x)$ $= f′(x)$

$= lim_{δ \to 0} \frac{f (x + δ) - f (x)}{δ}$ $= lim_(delta->0) (f(x+delta) - f(x))/delta$

Outline

The outline of material to learn "Differential Calculus" is as follows.

• Detailed outline of Differential Calculus

→ Application Scenario

→ Differentiation in First Principles

→ Graphical Meaning of Differentiation

→ Differntiability

→ Algebra of Derivatives

→ Standard Results