Admiral Payne 18:44, 25 October 2005 (UTC)
The basic law of differentiating functions is:
f(x)=x^n becomes f'(x)=nx^(n-1)
Below, I will try to prove this law, in several (I hope, easy) steps.
In order to differentiate it, you have to understand what differentiating is.
Differentiating is the differentiation quotient in one point.
Basic differentiation quotient calculation is basically "what's the difference in y-value between these x-values?"
When we want to get to differentiating, we just have to make a differentiation quotient for one point.
This would lead to a frequently used -although non sophisticated way- of making this differentiation function:
(f(x+0.0001)-f(x))/0.0001 (of course you could add more zero's, but it's kindof silly anyway, so forget about it)
What's more interesting though, because we want to find proof to the law, we seek something that always goes.
So, let's just forget about them small numbers and make it variable, so we can play with it:
(f(x+dx)-f(x))/dx the more delta x (dx) approaches zero, the closer it will get to differentiated function of f(x)
In order to use this function to prove our law, we're going to use f(x)=x^n on this function:
((x+dx)^n-x^n)/dx
Now, if you get the x+dx out of the brackets, it would lead to something among the lines of:
(x^n+nx^(n-1)*dx+nx^(n-2)*dx^2+ ... +nx^2*dx^(n-2)+nx*dx^(n-1)+dx^n-x^n)/dx
Now then, if we look carefully, we see that x^n and -x^n cancel eachother out, so it becomes:
(nx^(n-1)*dx+nx^(n-2)*dx^2+ ... +nx^2*dx^(n-2)+nx*dx^(n-1)+dx^n)/dx
We can even work out more things out of this big sum. We see that the dx appears in alot of states, so let's get some out!
nx^(n-1)+nx^(n-2)*dx+ ... +nx^2*dx^(n-1)+nx*dx^n
Right, now we'll just make the delta x (the difference in x) go to zero, this would lead to our proof!
Explaination to statement: This would mean that all the dx variables would be 0 (because 0*x=0):
So, that would basically mean that:
nx^(n-1)+nx^(n-2)*0+ ... +nx^2*0^(n-1)+nx*0
Of course, we again here that 0*x = 0 (which is of course, plain logic).
So, this would lead to our well-proven law!
f'(x)=nx^(n-1)