If you have ever came across derivative problems, the difficult questions and simple concept of differentiation obviously leaves you perplexed and impotent student. Some teachers teach differentiation in somehow complicated manner. Well, dealing with differentiation problems and conceptual questions will be a new nadir in your life just as like ** Algebra and Trigonometry.** After thorough practice, we find that nothing is difficult except by the way you approach or manipulate the thing.

So, let’s begin with differentiation as of now, forget everything and be ready to grasp the entire concept.

The slope of the secant is the average rate of change which measures always *‘the approximate rate of change in phenomena’*

OR

‘*Average rate of change y per unit change in x’ is known as derivative.*

The average rate of change or derivative can be calculated from the equation:

Let’s do an example on this,

**Determine the average rate of change of ‘y’ per unit change in ‘x’ for y = x² – 6x + 5 as ‘x’ increases from x = 1 to x = 3.**

**Given:** f(x) = y = x² – 6x + 5

x₁ = 1

x₂ = 3

Δx = 2

Plugging all the given values in the * ‘derivative formula’* yields:

We got * 2x + Δx – 6* as the derivative, plug in the given values in the derivative to get the desired value of instantaneous rate of change or derivative of the following question.

*= 2x + Δx – 6*

= 2(1) + 2 – 6 = -2

Easy? Let’s continue and move ahead for something easier than this. Suppose, we need to find out the average rate of change or derivative of every question. For this, you must learn some simple tips and formulas on how to evaluate every question. With this average rate of change formula, we can can also try other examples as well.

**Definiiton of derivative of a function as an instantaneous rate of change of a variable with respect to another variable**

We discussed that the average rate of change is the slope of the secant line joining the two points on the curve ** y =f(x).** More precisely, we are asked to determine the exact or instantaneous rate of change at a particular time. For example, for an airplane, what is the instantaneous rate of change of the distance that occurs at a specific time? or what is its instantaneous rate of change of the velocity at a specific time? This can be dealt by the slope of a tangent line to a curve y = f(x) at a particular point.

To illustrate this idea, let us examine the graph of a function ** y = x^{2}** at a particular point

**P (0.5, 0.25)**with different secant lines

**that is developed from the secant line**

*PQ*_{1}, PQ_{2}, …

*PQ.*The tabular form contains coordinates for the poitns ** P, Q,** the change

**x in**

*Δ***‘x’**, the change

**y in**

*Δ***‘y’,**and

**the slope of the secant lines**

*Δy /Δx,***Notice, that the slope of the secant line PQ is 2.5 (**

*PQ, PQ*_{1}, PQ_{2}, …

*Δy/ Δx***= 3.75/1.5 = 2.5**). If we take values of

**Q**closer to

**P**(i.e to Q

_{1}. Q

_{2}, Q

_{3}, …), then,

**x gets smaller and smaller and finally tends to zero.**

*Δ*The tabular form clearly shows that as **Q** approaches *P,* ** Δx** approaches 0 and the slope of the secant line obviously approaches the slope of the tangent line at a particular point

**which is**

*P (0.5, 0.25)***1.**This value

**i.e 1**is the derivative of the function

*y = f(x)*Geometrically, the slope of the tangent line to a curve at a particular point **P** is the instantaneous (or exact) rate of change at that particular point. This terminology develops the idea that the slope of the secant line becomes a better approximation for the slope of the tangent line to the curve at a particular point **P.**

From our discussion on limit, it follows that the exact/actual slope of the tangent line to a curve ** y = f(x)** at a particular point

**P**corresponds to the instantaneous rate of change at that specific point

**Find out the average rate of change of the following functions over the indicated intervals:**

**Q1. y = x^2 + 4 from x = 2 to x = 3**

**Q2. h = 2t-7 from t = 8 to t = 8.5**