(Y2-Y1) / (X2-X1)
This formula is another way of multiplying the height of the tangent by the base.
Looking at these results, we start to see a pattern. For the line Y=X^2, you just times the X co-ordinate by 2. For the Y=X^3, you times the X co-ordinate by 3, then square it. Again, for the Y=X^4 line, you times the X co-ordinate by 4, then cube it.
We can write these out as equations:
Y=X^2 (the gradient of the tangent equals) 2X^1
Y=X^3 (the gradient of the tangent equals) 3X^2
Y=X^4 (the gradient of the tangent equals) 4X^3
There is a definite pattern in the equations.
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It can also be written as Y=X^n (the gradient of the tangent equals) nX^(n-1)
To prove that this pattern works on all curves, I will use the same rule for the lines:
Y=X^5
Y=X^6
Y=X^7
We also see that these results are more accurate than the first set of results. But they are still not as accurate as they could be. To get an even more accurate reading, we need to make the increments even smaller. I will now use an increment of 0.000001.
Hopefully this will give an accurate enough reading. I will also work out, using the formula Y=X^n (the gradient of the tangent equals) nX^(n-1) to find out the actual gradient at each point on each curve. This will show how close my answers come to them.
Y=X^5
Y=X^6
Y=X^7
As you can see, the answers I got are very close to the actual answers. We still get a slight inaccuracy, as most calculators round up/down during the process, which can cause slight differences. In general though, these are very close and prove that the formula Y=X^n (the gradient of the tangent equals) nX^(n-1) works. This rule will work for any curve of the form Y=X^n with positive integer powers. But what I don’t know is if it works with fractional and negative powers.
I will now use the same methods and see if the rule Y=X^n (the gradient of the tangent equals) nX^(n-1) works with fractional powers.
Y=X^1/2
Y=X^1/4
Y=X^1/6
As you can see, using the same methods of calculation and the same formulas, the formula Y=X^n (the gradient of the tangent equals) nX^(n-1) does work with fractional powers. The answers are very close together showing a good accuracy.
I now need to see if the formula works with negative powers:
Eg Y=X^-1
I have drawn the graphs for Y=X^-1 and Y=X^-2 to show how the curve goes down instead of going up like the other curves. This shows the tangent has a negative gradient whereas the previous graphs have a positive gradient.
Y=X^-1
Y=X^-2
Y=X^-3
Again, you can see that the rule works for negative powers as well. With an increment this small, the gradients come out very close to what they should. This gives a greater scale of accuracy. As I predicted, all the gradients of the tangents are negative.
Overall, using the technique of drawing the tangents onto the graph, then dividing the height by the base, I found that it gave me a rough estimate of the gradient. However, I found this very inaccurate. Instead, I used the form (Y2-Y1) / (X2-X1) to give a more accurate answer using small increments on the inside of the curve. To go even more accurate, I used a very small increment of 0.000001. This was very accurate.
Using the answers I got, I then found a rule, Y=X^n (the gradient of the tangent equals) nX^(n-1). I found that this rule worked for any curve of the form Y=X^n where n equals a positive integer. I discovered that this rule was the first principle of differentiation.
To go even further, I decided to investigate if this rule worked with any other forms of n. I used the same techniques of small increments and found that the rule also worked with negative integers and fractional powers. So n can equal any number, and the rule will work. I also found that with a negative power, the gradients of the tangents would always equal a negative number.
I found a general rule that will work with any curve of the form Y=X^n even if n equals a positive, negative or fractional number.
- Y=X^n (the gradient of the tangent equals) nX^(n-1)
