Calculus: the Derivative Formula
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parametric differen.
equation of a tangent
equation of a normal
rate of change prob.1
rate of change prob.2
the Chain Rule
Chain Rule probs. #1
Chain Rule probs. #2
Chain Rule probs. #3
the Product Rule
parametric eqs.prob#1
parametric eqs.prob#2
intro. to integration
integration by parts 1
integration by parts 2
area under a curve
volumes of revolution
area between curves
Binomial Theorem
Bin. Theorem problems
Trig. Identities
Half Angle Formula
Double Angle Formula
Vector problems
Vector problems in 3D
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derivative formula
tangents & normals
inverse functions
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Edxl C1 Pure specimen
Edxl C1 Pure answers
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the derivative formula
tangents & normals
maxima & minima
chain rule
diffn.trigonometric fns.
product rule
quotient rule
parametric equations
implicit equations
differential equations
integration formula
int.'by substitution'
int.'by parts'
algebraic fractions
definite integrals
areas under curves
volumes of revolution
Trapezium Rule
integ. diff. equations
sine, cosine, tangent
sec, cosec, cotan
Sine & Cosine Rules
Pythagorean ID's
compound angles
Sigma Notation
arithmetic progression
geometic progression
line between points
more straight lines
parametric equations
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[ 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:

gradient of a straight line

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 :

gradient of straight line#2

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.

lim gradient


y = f(x)

y + dy = f(x + dx)

we have:

gradient f(x)

that is,



Example: find the gradient of  y = 4x2


gradt y=4x^3

gradient example expanded

gradient expanded

gradt. terms collected

cancelling by dx expression

in the limit when dx = 0 this becomes,

gradt. in the limit

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, then

the 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,

dervation with k

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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