Differentiating Functions: Difference between revisions

Revision as of 00:23, 28 October 2005

Revision as of 06:05, 28 October 2005

Line 2:

The basic law of differentiating functions is:

<math>f(x)=x^n</math> becomes <math>f'(x)=nx^({n-1)}</math>

Below, I will try to prove this law, in several (I hope, easy) steps.

Line 17:

This would lead to a frequently used -although non sophisticated way- of making this differentiation function:

(<math>\frac{f(x+0.0001)-f(x))/}{0.0001}</math> (of course you could add more zero's, but it's kind of 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.

Line 23:

So, let's just forget about them small numbers and make it variable, so we can play with it:

(<math>\frac{f(x+dx)-f(x))}{dx}</dxmath> 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:

(<math>\frac{(x+dx)^n-x^n)}{dx}</dxmath>

Now, if you get the x+dx out of the brackets, it would lead to something among the lines of:

(<math>\frac{x^n+nx^(n-1)*dx+nx^(n-2)*dx^2+ ... +nx^2*dx^(n-2)+nx*dxx^(n-1)+}{dx^n-x^n)}</dxmath>

Now then, if we look carefully, we see that x^n and -x^n cancel eachother out, so it becomes:

(<math>\frac{nx^(n-1)*dx+nx^(n-2)*dx^2+ ... +nx^2*}{dx^(n-2)+nx*dx^(n-1)+dx^n)}</dxmath>

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!

<math>nx^(n-1)+nx^(n-2)*dx+ ... +nx^2*dx^(n-1)+nx*dx^n</math>

Right, now we'll just make the delta x (the difference in x) go to zero, this would lead to our proof!

Line 47:

So, that would basically mean that:

<math>nx^(n-1)+nx^(n-2)*0+ ... +nx^2*0^(n-1)+nx*0</math>

Of course, we again here that 0*x = 0 (which is of course, plain logic).

Line 53:

So, this would lead to our well-proven law!

<math>f'(x)=nx^(n-1)</math>