

First Principles
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 Q_{1} Q_{2} Q_{3} Q_{4} Q_{5} Q_{6} ...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.
We know from coordinate geometry that:
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 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.
If we now replace y by f(x) in the expression for gradient, since y = f(x) i.e. y is a function of x.
and
y = f(x)
y + dy = f(x + dx)
we have:
that is,
Example: find the gradient of y = 4x^{2}
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.
Notation This is best described with an example.
If y = 3x^{2} , which can also be expressed as f(x)= 3x^{2}, then
the derivative of y with respect to x can be expressed as:
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 x^{2} at the point (2,1).
gradient = (5) (2 x^{21}) = 10 x^{1} = 10 x
gradient at point (2,1) is 10 x 2 = 20