To begin explaining why differentiation works we have to consider what the gradient of something actually is, and that is the rate of change of a graph. The rate of change is the: change in y/change in x (which is what dy/dx means). So if we have some function, f(x) and we want to find the gradient of it at any given point it may be sensible to think that to find the gradient of a curve at any point (x, f(x)) we do:

But, this means that the denominator is 0, so the gradient of the curve according to this is undefined, which makes no sense at all. So we need to think of another way to try and find the gradient of a curve.

Let us consider the graph y = 2x

^{2}+ 3x - 1, if we want to find the general gradient for any point x, let us consider the line created by the two x co-ordinates x and x + h, the y co-ordinates of these points are 2x

^{2}+ 3x - 1 and 2x

^{2}+ 4xh + 3x + 2h

^{2}+ 3h - 1 respectively. So the gradient of this line is:

So this says that the gradient of y = 2x

^{2}+ 3x - 1 is 4x + 3, which agrees with differentiation as we know it. But this is just an example that differentiation works, not a proof that it is true for all values. But, applying the same principles we used in our example we can prove it for all values.

Let f(x) = ax

^{n}, consider the line from connecting two points x and x + h, the y co-ordinates for these points is ax^{n }and a(x+h)^{n}, the expansion for this may not seem immediately apparent but if you have encountered Pascal's triangle and the binomial expansion you will understand that a(x+h)^{n }can be expanded to an extent. To fully understand what I will be next doing you may want to read up on the binomial expansion. If you really do not understand this next step, leave a comment and I will explain to the best of my ability.
And that is it! Hopefully you managed to understand what I have done, but if any points have you confused, please leave a comment. If you liked this, why not like us on Facebook? Go on, you know you want to.

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