Once again, this did not fit a pattern that I was expecting as they were not 4 multiplied by x or x4. From the three graphs that I drew, I concluded that it was necessary for me to find another way of finding a formula for the gradient as there was clearly no simple formula for all graphs that could be seen just from looking at the table of values.
The Small Change Method for Calculating Gradients and Limiting Values
This method allows you to calculate the gradient of a point on the curve by drawing chords that become shorter and shorter, as shown in the diagram below.
The small change theory makes it possible to calculate the gradient fairly accurately. The bigger the chord (AB) the less accurate the gradient will be as it is the furthest distance from the curve. Thus, the shorter the chord, the more accurate it will be. The small change method uses the original formula for gradient (δy/δx) but places the second point very close to the first one as shown in the diagram on the right.
By putting the values in the diagram into the formula, I would be able to work out a gradient that was precise. Hopefully, my calculated gradient should be very close to 4.
Gradient = δy
δx
Gradient = 2.0012 – 4
2.001 – 2
Gradient = 4.001
Further Investigation of the Small Change Theory on more complex graphs
I then tried this for the point (3,27) on my x3 graph. I expected the gradient to be somewhere about 28.
Gradient = δy
δx
Gradient = 3.0013 – 33
3.001 – 3
Gradient = 27.001
Generalising the small change theory and proving the formula for the gradient of y =x2
This method allows me to calculate the gradient numerically but this does not show how the formula for the graph of y = x2 is 2x. Therefore, I decided to prove the small change method for y = x2 in more general terms, as is shown in the diagram on the left. I have replaced the small change of 0.001 by the letter d and the coordinates of (2,4) by x, x2. Therefore, the second set of coordinates must be replaced by (x+d)2 (x+d).
I then decided to calculate the gradient of these coordinates using the same formula for the gradient:
Gradient = δy
δx
Gradient = (x+d)2 – x2
(x+d) – x
Gradient = x2 + 2xd + d2
x + d – x
I cancel out the x2s and the xs so:
Gradient = 2xd + d2
d
I then factorised the top of the equation by d so that I could cancel out the ds
Gradient = d(2x + d)
d
Therefore the gradient of the graph y = x2 is 2x + d.
At first, I thought that this must be wrong because it was not simply 2x. However, I realised that d was so small that in the context of finding the gradient, it was almost negligible. Hence, to calculate the gradient normally it is simply Gradient = 2x.
* * *
I was not able to find a formula for the graph x3, so I decided to use the above method to find out what it was.
Gradient = (x + d)3 –x3
x + d – x
I expanded the brackets using Pascal’s Triangle and the Binomial Theorem which will be mentioned after I have found the formula for y = x3.
Gradient = x3 + 3x2d + 3xd2 + d3 – x3
x + d – x
As with the previous example I can cancel out the x3s and the xs. Therefore:
Gradient = 3x2d + 3xd2 + d3
d
All of the expressions on the top have d in, so I can factorise the top half of the equation:
Gradient = d(3x2 + 3xd + d2)
d
Thus, having cancelled out the ds (as shown):
the formula for the gradient of y = x3 is 3x2 + 3xd + d2
As before, the only term that we use in most situations is the 3x2 as it does not contain a d which is so close to zero that when x is multiplied by it, the answer is so tiny that it is negligible.
Pascal’s Triangle and The Binomial Theorem (I)
The above diagram is Pascal’s Triangle for values from 1 – 10. We use Pascal’s Triangle to help us expand brackets which are to the power of a number. This definitely works for all positive integers and later on in the investigation, I shall look at whether this is true for square roots, negative numbers and decimals.
I shall demonstrate the use of Pascal’s Triangle by expanding the brackets (x + d)4
1 4 6 4 1
The line of Pascal’s Triangle which has a four in is the one we need to use. We take the 1 which is the coefficient of the first term, x4 . We then have a 4 which gives us the coefficient of x in the second term. From the Binomial Theorem, I know that the powers of x decrease by one as you go through the equation, and the powers of d increase by one.
Therefore:
x4 + 4x3d + 6x2d2 + 4xd3 + d4
This can be done for any number and is useful when expanding brackets which are to high powers.
General Formula For The Gradient of Positive Graphs
I used the above method to expand different brackets from (x+d)2 to (x+d)7. In the table below, I have only included the relevant term
From this table I can see a pattern – that the coefficient of x is the same as the original power and the power has decreased by one. Therefore the formula for finding the gradient of a curve at a particular point (x) is:
nx(n-1)
Proving the General Formula
To prove the general formula, it was necessary to expand the brackets (x + d)n. This proved quite difficult and several sources of information were used including books on the subject and the Internet.
When I expand the brackets and substitute the answer into the formula of δy/δx, I have managed to prove that it is indeed nx(n-1). The way in which I have done this is shown below:
Gradient = δy
δx
δy = xn + nx(n-1)d + n(n – 1) x(n-2)d2 + ....n(n-1)...(n – r +1) x(n-r)dr + dn - xn
δx 2! r!
x + d – x
δy = nx(n-1)d + n(n – 1) x(n-2)d2 + ....n(n-1)...(n – r +1) x(n-r)dr + dn
δx 2! r!
d
δy = d(nx(n-1) + n(n-1) x(n-2)d + .... n(n-1)...(n-r + 1) x(n-r)d(r-1) + d(n-1))
δx 2! r!
d
δy = nx(n-1) + n(n-1) x(n-2)d + .... n(n-1)...(n-r + 1) x(n-r)d(r-1) + d(n-1)
δx 2! r!
When we consider, as we have done for all other proofs of formulas, that d is so small that it is negligible and effectively cancels out all terms that it is multiplied by. Thus, we are left with simply nx(n-1) which is the general formula for finding the gradient and also the formula for differentiation which is the simple way of calculating the gradient.
It is also true that the formula for (x + d)n can be written as:
n
(d+x)n ≡ Σ n dn-rxr
r=0 r
This notation means exactly the same thing, the Σ standing for the addition of all the terms in the series and the n and r = 0 standing for the limits of the equation. This is a shorter way of writing the formula that I have deduced above and I found it in a mathematical textbook.
Further Investigation Into Curves of My Choice
For this section of the investigation, I intend to look at trigonometrical curves. I know that I shall not be able to differentiate out sin x using simply nx(n-1) as it will not work. I am also not able to fully understand why sin x is cos x so I am simply demonstrating how this function works rather than trying to explain it.
This type of curve is probably one of the most difficult types to calculate the gradient. For this I shall be looking at the most ‘simple’ of the trigonometrical graphs – y = sinx. I know, from having read up on the subject, that when we differentiate y = sinx, the answer is cosx, very different to the formula nx(n-1) that I discovered for positive graphs. Below, I hope to show that y = sinx is indeed cos x .
To do trigonometrical differentiation, it is necessary to do all calculations in radians, otherwise it does not work. Although I do not fully understand this concept, I have gathered that it is to do with the rate of change along the x axis.
Thus:
δ sinx = sin(x + δx) – sin x
δx δx
δ sin 0.5 = sin (0.5 + 0.0001) – sin 0.5
δx 0.0001
= sin (0.5001) – sin 0.5
0.0001
= 0.000087755
0.0001
= 0.87755859
If, still working in radians, we do the calculation of cos 0.5, the answer that we are given is 0.877582561. This is extremely close to the answer that we get for differentiating sin 0.5 and therefore the derivative of sin x is cos x. However, I cannot show, apart from the algebra above, or explain why this works, although I have tried.
The second trigonometrical curve that I shall investigate is y = cos x. I know from a textbook that the derivative of this is y = -sin x and I hope to prove that this is true. As we are dealing with trigonometrical function we still need to work in radians. Hence y = cos 0.5 is y = -sin 0.5
δ cos 0.5 = cos (0.5 +0.00001) – cos 0.5
δx 0.00001
= cos(0.50001) – cos 0.5
0.00001
= - 0.000004794
0.00001
= -0.4794298
Conclusion
Having investigated the gradient function for graphs of the type xn, I have come to the conclusion that the formula nx(n-1) is true for all types that I have investigated except for the trigonometrical graphs. I have proved that this is true graphically, numerically and algebraically. I have looked at tables of values, found formulas for each type of graph and then proved that these formulas are true, using algebra.
I have furthered my investigation by looking at the gradient of trigonometrical graphs and shown that the theory that I found in a textbook is true. Naturally, this investigation could be expanded further and further, looking at all types of curves such as negative curves and those to the power of different numbers.
I have completed the objectives that I set myself at the beginning of the investiagtion. I am now able to calculate any gradient accurately of any point on almost any graph without drawing a tangent. I have also completed the objective of proving my formulas and generalising the rule. Thus to find the gradient the formula is
nx(n-1)
