I will now show the gradients of chords starting at point (2,4) and finishing at various other points along the plotted line ie. 2 to 5 and 2 to 4 etc.
You can see the pattern that is emerging.
I shall attempt to prove this by using a formula that is provided in the ‘Essential Pure Mathematics’ A-level textbook.
The formula is:-
RQ = NQ – NR
= (x + h)² - x²
= 2xh + h²
and
PR = h
The gradient of the chord PQ is:-
RQ = 2xh + h²
PR h
= h(2x + h)
h
= 2x + h
As I rotate the chord clockwise about point P, Q approaches P along the curve and the gradient of the chord PQ is the gradient of the tangent at P, and h will be 0.
So when h will be 0, the gradient of the PQ chord is 2x + h, so when h = 0 the formula will be gradient = 2x.
As x² is an expression where we substitute a value for x to find the y-coordinate to plot the graph y = x², we now have another expression, 2x, in this we will substitute the value of x to find the gradient of points along the graph.
2x is the gradient function of the curve y = x².
This can also be written as grad x² = 2x.
Here are some examples for when y = x²:-
When x = 5 y = 5² = 25
The gradient function is 2x.
The gradient = 10 when x = 5
The point is (5,25) and the gradient is 10.
I have checked this result and it is correct.
will now investigate the gradient of y = x³. A graph to show this is below:-
I have found out the gradient for each point on the graph. I did this in the way that I explained on page1. A table shows this below:-
Diff 1 3 9 15 21 27
Diff 2 6 6 6 6
This means that my formula for the gradient is:-
3x²
Here is a table for the chords:-
Diff 1 5.75 5.25 4.75 4.25 3.75
Diff 1 0.69 0.67 0.65 0.63
As you can see there is a pattern emerging from the chords of these numbers.
Below is another copy of the graph y = x³.
I will now attempt to prove my formula which is:- (I have again used ‘Essential Pure Mathematics’ A-level textbook)
Gradient = 3x²
RQ = NQ – NR
= (x + h)³ - x³
= x³ + 3x³h +xah² + h³ - x³
= 3x²h + 3xh² + h³
PR = h
The gradient of PQ = RQ
PR
= 3x²h + 3xh² + h³
h
= 3x² + 3xh + h²
As Q moves closer to P along the curve, h will be 0. So then the terms 3xh and h² will both be zero. Therefore the gradient of PQ = 3x².
So grad x³ = 3x².
Here are some examples for when y = x³:-
When x = 3 y = 3³ = 9
The gradient = 6, when x = 3
The point is (3,9) and the gradient is 6.
I have checked this result and it is correct.
I now have confirmed the following formulae:-
Grad x² = 2x
Grad x³ = 3x²
I will now predict that:-
Grad x = 4x³
My proven results suggest that the rule for differentiating a power of x is:-
Multiply by the index and reduce the index by 1
This means that:-
Grad x = 5x
Grad x = 6x
So my general formula will be:-
Grad x = nx
I know from previous class investigations into straight line graphs ie. y = 6 and y = -3 that these graphs have a gradient of zero. So Grad 4 = 0 and Grad –3 = 0.
So if we differentiate a constant we get 0.
So far this agrees with my general formula as x ° =1. We can write this as Grad 4 = Grad 4x ° = 0 × 4x = 0.
Again I know that graph y = mx + c has a gradient of m and graph y = x has a gradient of 1 and graph y = 6x has a gradient of 6.
So Grad x = 1
Again this agrees with my formula as:-
Grad x = 1 × x ° = 1
and,
Grad 6x = 6 × grad x = 6 × 1 = 6.
This shows the general fact that if a function has a factor which is constant, that constant remains the same as a factor of the gradient function.
This concludes my coursework.
