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Thought-Process to Discover Knowledge

Welcome to nubtrek.

Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.

In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.

Read in the blogs more about the unique learning experience at nubtrek.
mathsDifferential CalculusIntroduction to Differential Calculus

### Differentiation: First Principles

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A car travels a distance 20m in 2 seconds, what is the speed of the car?

• 20 m/sec
• 10 m/sec
• 10 m/sec

The answer is "10 m/sec"

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

Speed is rate of change of distance.

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

• without a numerical value for t the distance traveled cannot be computed.
• without a numerical value for t the distance traveled cannot be computed.
• For any value of t, the distance traveled is s=20t meter

The answer is "For any value of t, the distance traveled is s=20t meter".

Measurement can be expressed as a function of a variable.

A car is moving with displacement given as a function of time s=3t^2+2. What is the distance at t=2sec ?

• 14 meter
• 14 meter
• 6 meter

The answer is "14 meter"

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

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

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

Which one is correct?

• Person A
• Person B
• Neither of them
• Neither of them

The answer is "Neither of them". The answer is explained in the subsequent questions.

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

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

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

What have person A and B calculated?

• Person A calculated average speed till 2sec
• Person A calculated average speed till 1sec
• both the above
• both the above

The answer is "both the above".

The results from these two are different and so the average speed changes with time.

A car is moving with displacement given as a function of time s=3t^2+2. Which of the following is expected?

• speed is a numerical value
• speed is a function of time
• speed is a function of time

The answer is "speed is a function of time"

A car is moving with displacement given as a function of time s=3t^2+2. Which of the following is correct?

• Speed is rate of change of displacement. So, one has to figure out how to find the rate of change of an algebraic expression.
• Speed is rate of change of displacement. So, one has to figure out how to find the rate of change of an algebraic expression.
• Speed is defined only when distance traveled in a time period is given. If the distance is given as a function of a variable, speed cannot be calculated.

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

A car is moving with displacement given as a function of time s=3t^2+2. Speed is the rate of change. Few students are set to find rate of change
students may work these out to understand

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

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

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

What is actually being calculated by each?

• Average of two values representing speed at t and t+delta
• Average of two values representing speed at t and t+delta
• instantaneous speed for any given time

The answer is "Average of two values representing speed at t and t+delta". The calculation does not provide instantaneous speed.

A car is moving with displacement given as a function of time s=3t^2+2. A speedometer is attached to a wheel. The speedometer measures the speed at which the wheel rotates and proportionally provides the speed of the car. What is the speed shown in the speedometer?

• Average speed over time
• instantaneous speed for any given time
• instantaneous speed for any given time

The answer is "instantaneous speed for any given time".

A car is moving with displacement given as a function of time s(t)=3t^2+2. A speedometer can be used to measure the instantaneous speed of the car.

The average of speeds at t and t + delta is (s(t+delta)-s(t))/delta =6t+3delta.

Can the instantaneous speed be computed as an algebraic expression from the given function?

• No. Instantaneous speed can only be measured.
• Yes. Instantaneous speed is when the time interval delta is zero.
• Yes. Instantaneous speed is when the time interval delta is zero.

The answer is "Yes. Instantaneous speed is when the time interval 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.

A car is moving with displacement given as a function of time s(t)=3t^2+2. The instantaneous speed is (s(t+delta)-s(t))/delta |_(delta=0). What is the value of this calculation when delta=0 is substituted?

• it is indeterminate value 0//0
• it is indeterminate value 0//0
• it is 0 as the numerator is zero

The answer is "it will be indeterminate value 0//0".

A car is moving with displacement given as a function of time s(t)=3t^2+2. The instantaneous speed is

(s(t+delta)-s(t))/delta |_(delta=0) = (s(t)-s(t))/delta =0/0

That is indeterminate value.
How does one solve a function evaluating to indeterminate value?

• Use Limit of the function as delta approaching 0.
• Use Limit of the function as delta approaching 0.
• There is no method to solve a function evaluating to 0//0

The answer is "Use Limit of the function as delta approaching 0."

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

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

lim_(delta->0) (color(deepskyblue)(s(t+delta))-color(coral)(s(t)))/delta

quad quad = lim_(delta->0)(color(deepskyblue)(3(t+delta)^2 + 2) - color(coral)(3t^2 -2) )/delta

canceling color(deepskyblue)(2) and color(coral)(-2) and expanding the square
quad quad = lim_(delta->0) (color(deepskyblue)(3t^2+6t delta + 3delta^2) -color(coral)(3t^2))/delta

canceling color(deepskyblue)(3t^2) and color(coral)(-3t^2)
quad quad = lim_(delta->0) color(deepskyblue)(6t delta + 3delta^2)/delta

canceling delta from numerator and denominator
quad quad =lim_(delta->0) color(deepskyblue)(6t + 3delta)

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

The instantaneous speed is computed as an algebraic expression.

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 delta->0 in the variable.

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

Differentiation in the context of cause-effect pair: If effect is given by f(x) then the cause is computed as differentiation or derivative of f(x) denoted as d/(dx) f(x) or f′(x).

d/(dx) f(x) = lim_(delta->0) (f(x+delta)-f(x))/delta

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

Note that (du)/(dx) is another quantity related to the given quantity u.

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

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

(dy)/(dx)

=y′

= lim_(Delta x ->0) (Delta y)/(Delta x)

=d/(dx) f(x)

= f′(x)

= lim_(delta->0) (f(x+delta) - f(x))/delta

= lim_(Delta x ->0) (f(x+Delta x) - f(x))/(Delta x)

What is rate of change of a function called?

• Pronunciation : Say the answer once

The answer is "differentiation or derivative of the function".

Students can connect the notation d/(dx) f(x) as

•  the small difference in x is given by denominator dx

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

•  d/(dx) denotes the difference in function with respect to a small difference in x.

The rate of change of a function with respect to the variable is the derivative of the function.

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

d/(dx) f(x)

= f′(x)

= lim_(delta->0) (f(x+delta) - f(x))/delta

Solved Exercise Problem:

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

(dy)/(dx)

=lim_(delta->0) [color(coral)(f(x+delta))- color(deepskyblue)(f(x))]//delta

=lim_(delta->0) [color(coral)((x+delta)^2+ x+ delta) - color(deepskyblue)((x^2+x))]//delta

=lim_(delta->0) [color(coral)(x^2+2delta x + delta^2 + x + delta) - color(deepskyblue)(x^2 -x)]//delta

=lim_(delta->0) [color(coral)(2delta x + delta^2 + delta)]//delta

=lim_(delta->0) 2x+delta+1

What does the above prove?

• d/(dx) (x^2+x) = 2x+1
• d/(dx) (x^2+x) = 2x+1
• d/(dx) (x^2+x) = 2x+1+ delta

The answer is "d/(dx) (x^2+x) = 2x+1"

Solved Exercise Problem:

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

(dy)/(dx)

=lim_(delta->0) [color(coral)(f(x+delta))- color(deepskyblue)(f(x))]//delta

=lim_(delta->0) [color(coral)(sin(x+delta)- color(deepskyblue)(sinx)]//delta

=lim_(delta->0) [color(coral)(sinx cos delta + cos x sin delta)-color(deepskyblue)(sinx)]//delta

= lim_(delta->0) [color(coral)(cos x sin delta)]/delta - lim_(delta->0) (color(deepskyblue)(sin x) - color(coral)(sin x cos delta))/delta

= cos x lim_(delta->0) (sin delta)/delta - sin x lim_(delta->0)(1 - cos delta)/delta

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

What does the above prove?

• d/(dx) sin x = cos x
• d/(dx) sin x = cos x
• d/(dx) sin x = cos x - sin x

The answer is "d/(dx) sin x = cos x"

slide-show version coming soon