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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.continue
mathsDifferential CalculusIntroduction to Differential Calculus

Differentiation: First Principles

In this page, differentiation is defined in first principles.



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A car travels a distance `20`m 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 `20`m/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=2`s, the distance traveled is `14`m. So the speed is `14/2 = 7`m/sec.

 •  Person B does the following: At `t=1`s, the distance traveled is `5`m. 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=2`s, the distance traveled is `14`m. So the speed is `14/2 = 7`m/sec.

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

What have person A and B calculated?

  • Person A calculated average speed till `2`sec
  • Person A calculated average speed till `1`sec
  • 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
    Spelling: Write 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`"

                            
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