In this post I intend to discuss the concept of differentiation in greater detail.In my last post we have already learned what derivatives are and why we need to learn this concept. Now let us try to find out exactly **what a derivative function tell us with respect to graphs.**

A Basic point we need to remember here, which I shall keep reiterating is –

**Rate of change = Slope of the line **

For many graphs, the steepness or the slope varies from point to point i.e some parts are very flat,some are less steep and some are very steep(see the graph below).This pattern indicates that the rate of change of the entity we are studying is NOT the same throughout.

- A positive derivative means that the function is increasing.
- A negative derivative means that the function is decreasing.
- Zero derivative means that the function has some special behavior at that given point.

- Consider the following graph-

We want to find out how much has the value of ‘y’ changed from point P to point T .This is called average rate change. Thus,

**Average rate change =slope of the secant line joining P&T = Δy/Δx =change in y/change in x**

(*Secant means a line that cuts a curve in two or more parts.Here line PQ & PS are secants.*)

Later suppose we want to find out the change between point P and point R.

**Average rate of change = Slope of secant line joining P & R. **

But, this slope will be different from the slope of the secant line joining points P and Q , as the steepness of both lines vary. So, basically the *average* *rate of change *(i.e slope) in both cases is NOT the same.

So we now try to find** instantaneous rate of change** (rate of change at a given instant ), which basically means we try to find the derivative of the function y with respect to x.

To understand the difference between *average rate of change* and *instantaneous rate of change* lets consider two cases –

- Case 1 –

x |
y |

0 |
5 |

5 |
10 |

10 |
15 |

15 |
20 |

Here, the average rate of change for first and second values = y2-y1/ x2-x1

= 15-5/5-0

= 10/5

= 2 (∴slope = 2)

Average rate of change for next set of values= y3-y2/x3-x2

= 35-15/ 10-5

= 20/5

= 4 (∴slope = 4)

Average rate change for next set = y4-y3/ x4- x3

= 0-35 /15-10

= -35/5

= -7 (negative slope)

Now here, the average rate of change is not constant , so we will **NOT** get a straight line on the graph. We expect to get a **curve** with different slopes at different points. So, finding average rate of change is not very helpful in this case. Thus, we switch to instantaneous rate of change concept.

**The instantaneous change can be represented by a tangent line on the graph**. Why only a tangent line? That’s because a tangent touches the curve roughly only at one point that is at that instant – so the slope of the tangent at that point represents the instantaneous change at that given point and we say that the function is

*differentiable*at that point.

But how does one get a tangent line from a secant line? If we gradually reduce the time interval i.e go on reducing the length of the secant line, as shown in fig below , at one point the secant becomes tangent at that point on the graph.

This basically means we are **dividing our data into small parts i.e differentiating** , so that we can go from *average rate change *phenomenon(which is not constant due to fluctuating data) to *instantaneous rate change *at a given point.We find rate changes at every point and later integrate everything to find the desired result.

So, we draw tangents at every desired point on the graph.Let’s see how that looks on the graph(fig 1) –

We then find the slope of these tangent lines.Why do we take slope?As previously stated,**slope represents rate of change **and then we plot these slopes separately as shown in fig 2. Guess what we get ?** **

**A STRAIGHT LINE**** !! BINGO! **And we already know how great it is to work with straight lines on a graph! So,we basically took derivative to get straight line on a graph,to simplify and better understand our experimental data !

*e.g.* While studying Freundlich adsorption isotherm(we shall study it in detail when I start writing about adsorption) we have to use this concept.The isotherm is expressed as:

x/m = kp^{1/n}

x/m ⇒ amount of gas adsorbed per unit mass of adsorbent

p ⇒ equilibrium pressure of the gas and

k,n ⇒ arbitrary constants.

Plot of this equation will not give straight line but after taking logs on both sides we get a straight line on the graph.

In the next post we shall continue our discussion on derivatives.Till then,

Be a perpetual student of life and keep learning !

Good day !

References and Further Reading –

1)http://www.maths.surrey.ac.uk/explore/vithyaspages/differential.html

2)http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/derivative/slope.html

3)http://www-math.mit.edu/~djk/calculus_beginners/chapter01/section02.html

4)https://www.math10.com/en/algebra/derivative-function.html

5)https://math.dartmouth.edu/opencalc2/cole/lecture8.pdf

7)http://www.modelab.ufes.br/xioste/papers/xioste_paper105.pdf