PURE MATHEMATICS - Differential Calculus

 

The Derivative Formula

 

 

first principles

notation

the formula

 

 

 

First Principles

 

 

chord tangent relation

 

 

To find an expression for the gradient of the tangent at point P on a curve, we must consider lines passing through P and cutting the curve at points Q1 Q2 Q3 Q4 Q5 Q6 ...etc.

 

As Q approaches P so the gradient of the chord PQ approaches the gradient of the tangent at P.

 

We can form an expression for the gradient at P by using this concept.

 

 

gradient of a straight line

 

 

We know from coordinate geometry that:

 

 

for points (x1y1) and (x2y2)

 

 

Consider the coordinates of P to be (x,y) and point Q to be (x+dx, y+dy), where dx and dy are the horizontal and vertical components of the line PQ.

 

 

gradient with dy and dx

 

 

Gradient of the line between points (x,y) and (x+dx, y+dy) is given by :

 

 

 

 

The tangent to the curve = gradient of PQ when the length of PQ is zero and dx = 0 and dy = 0.

 

In the limit, as dx 'approaches zero' the gradient of the curve is said to be dy/dx.

 

 

 

 

Since,

 

if we incrementally increase both y and x , then

 

 

 

 

Now we can substitute for y + dy in the top bracket of the 'gradient' equation.

 

At the same time we can replace the y by f(x) .

 

In the bottom brackets the +x and the -x cancel out, leaving dx

 

Then we have:

 

 

 

that is,

 

 

 

 

 

Example: Find the gradient of   y = 4x2.

 

 

 

 

cancelling by dx,

 

 

in the limit when dx = 0 this becomes,

 

 

 

 

Without doubt this is a very long winded way to work out gradients. There is a simpler way, by using the Derivative Formula.

 

 

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Notation This is best described with an example.

 

If  y = 3x2 , which can also be expressed as f(x)= 3x2, thenthe derivative of y with respect to x can be expressed as:

 

 

 

 

 

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The Derivation Formula

 

If we have a function of the type  y = k x n , where k is a

constant, then,

 

 

 

example #1

 

Find the gradient to the curve  y = 5 x2 at the point (2,1).

 

gradient = (5) (2 x2-1) = 10 x1 = 10 x

 

gradient at point (2,1) is 10 x 2 = 20

 

 

 

 

 

 

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