# Differentiation From First Principle

Contents

### Summary

4 steps to work out differentiation from the First Principle:

1. Give increments to both x & y i.e $\Delta x,\quad \Delta y$.
2. Find change $\Delta y$ of y.
3. Find rate of change of y with respect to x i.e  $\frac { \Delta y }{ \Delta x }$  or  $\frac { dy }{ dx }$.
4. Take the limit of  $\frac { \Delta y }{ \Delta x }$  as $\Delta x\quad \rightarrow \quad 0$.

We know that the gradient of a line is always constant and can be found from the following equation:

$Gradient\quad =\quad \frac { { y }_{ 2 }\quad -\quad { y }_{ 1 } }{ { x }_{ 2 }\quad -\quad { x }_{ 1 } }$

$=\quad \frac { difference\quad in\quad y\quad coordinate }{ difference\quad in\quad x\quad coordinate }$

A curve is not a straight line, hence its gradient is still not constant. It is changing from one point to another point on the curve. So we don’t have a set gradient for a curve.

Consider two points $P({ x }_{ 1 },\quad { y }_{ 1 })$ and $Q({ x }_{ 2 },\quad { y }_{ 2 })$ on a curve in ‘Fig 1’.

The gradient of the straight line:

$PQ\quad =\quad \frac { { y }_{ 2 }\quad -\quad { y }_{ 1 } }{ { x }_{ 2 }\quad -\quad { x }_{ 1 } }$

If we move from point Q towards point P, we are actually moving from ${ x }_{ 2 }$ towards ${ x }_{ 1 }$ and from $y_{ 2 }$ towards $y_{ 1 }$.

Hence when Q approaches , the line PQ becomes the tangent line at P.

We can say that now the gradient of line PQ is ‘m’ which is also known as the gradient of the curve at point P. Also ‘m’ is the gradient of the tangent at P.

Therefore, we define the gradient of a curve at a point P to be the gradient of the tangent drawn at that point.

We now find the gradient of a curve by the method known as ‘differentiation from the first principle’.

#### What is differentiation from the first principle?

To start of with, consider a curve with equation $y\quad =\quad { x }^{ 2 }$. Let P and Q be two points on the curve with coordinates P(x, y) and $Q(x\quad +\quad \Delta x,\quad y\quad +\quad \Delta y)$ where $\Delta x$ and $\Delta y$ represent small increments in x and y respectively.

For:        $y\quad =\quad { x }^{ 2 }$

Step 1:

$y\quad +\quad \Delta y\quad =\quad { (x\quad +\quad \Delta x) }^{ 2 }$

$y\quad +\quad \Delta y\quad =\quad { x }^{ 2 }\quad +\quad { (\Delta x) }^{ 2 }\quad +\quad 2x(\Delta x)$

Step 2:

Make $\Delta y$ the subject

$\Delta y\quad =\quad { x }^{ 2 }\quad +\quad { (\Delta x) }^{ 2 }\quad +\quad 2x(\Delta x)\quad -\quad y$

Substitute $y\quad =\quad { x }^{ 2 }$  in the above equation

$\Delta y\quad =\quad { x }^{ 2 }\quad +\quad { (\Delta x) }^{ 2 }\quad +\quad 2x(\Delta x)\quad -\quad { x }^{ 2 }$

$\Delta y\quad =\quad { (\Delta x) }^{ 2 }\quad +\quad 2x(\Delta x)$

As $\Delta x$ is common we take it to the other side of the equation.

Step 3:

$\frac { \Delta y }{ \Delta x } \quad =\quad \Delta x\quad +\quad 2x$

As Q approaches P $\Delta x$ becomes smaller and smaller and eventually becomes zero. At this instance, the gradient of PQ becomes the gradient of the tangent at P.

Step 4:

As   $Q\quad \rightarrow \quad P$,  gradient of  $QP\quad =\quad \lim _{ \Delta x\rightarrow 0 }{ (\frac { \Delta y }{ \Delta x } ) }$

$=\quad \lim _{ \Delta x\rightarrow 0 }{ (\Delta x\quad +\quad 2x) }$

$=\quad 0\quad +\quad 2x$

$=\quad 2x$

$\lim _{ \Delta x\rightarrow 0 }{ (\frac { \Delta y }{ \Delta x } ) }$  is called the differential coefficient of y with respect to x OR derivative of y with respect to x and is symbolically written as $\frac { dy }{ dx }$.

The above method of finding the differential coefficient of y with respect to x is known as “Differentiation from the First Principles”.

It involves four steps:

 Step 1 Give increments to both x & y i.e $\Delta x,\quad \Delta y$$\Delta x,\quad \Delta y$ Step 2 Find change $\Delta y$$\Delta y$ of y Step 3 Find rate of change of y with respect to x i.e $\frac { \Delta y }{ \Delta x } \quad or\quad \frac { dy }{ dx }$$\frac { \Delta y }{ \Delta x } \quad or\quad \frac { dy }{ dx }$ Step 4 Take the limit of $\frac { \Delta y }{ \Delta x } \quad as\quad \Delta x\quad \rightarrow \quad 0$$\frac { \Delta y }{ \Delta x } \quad as\quad \Delta x\quad \rightarrow \quad 0$

#### Example #1

Q. Differentiate $\frac { 2 }{ x }$ with respect to x from the First Principle.

Firstly, let    $y\quad =\quad \frac { 2 }{ x } \quad \quad \quad \quad \quad \rightarrow \quad equation\quad 1$

Step 1:

Giving increments to x & y

$y\quad +\quad \Delta y\quad =\quad \frac { 2 }{ x\quad +\quad \Delta x }$

Step 2:

$\Delta y\quad =\quad \frac { 2 }{ x\quad +\quad \Delta x } \quad -\quad y$

Substitute equation 1 in the above equation

$\Delta y\quad =\quad \frac { 2 }{ x\quad +\quad \Delta x } \quad -\quad \frac { 2 }{ x }$

$\quad =\quad \frac { 2x\quad -2x\quad -2\Delta x }{ x(x\quad +\quad \Delta x) }$

$\quad =\quad \frac { -2\Delta x }{ x(x\quad +\quad \Delta x) }$

Step 3:

Divide both sides by $\Delta x$

$\frac { \Delta y }{ \Delta x } \quad =\quad \frac { -2\Delta x }{ x(x\quad +\quad \Delta x)\Delta x }$

$\frac { \Delta y }{ \Delta x } \quad =\quad \frac { -2 }{ x(x\quad +\quad \Delta x) }$

Step 4:

When   $\Delta x\quad \rightarrow \quad 0$

$\lim _{ \Delta x\rightarrow 0 }{ \frac { \Delta y }{ \Delta x } } \quad =\quad \lim _{ \Delta x\rightarrow 0 }{ (\frac { -2 }{ x(x\quad +\quad \Delta x) } ) }$

$=\quad (\frac { -2 }{ x(x\quad +\quad 0) } )$

Ans:        $\frac { \Delta y }{ \Delta x } \quad =\quad \frac { -2 }{ { x }^{ 2 } }$