Differentiating Functions: Difference between revisions

Revision as of 06:15, 28 October 2005

Revision as of 06:18, 28 October 2005

Line 31:

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

<math>lim_{h\rarr0}\frac{x^n+nx^{n-1}h+nx\frac{1}{2}n(n-1)x^{n-2}h^2+ ... -x^n}{h}</math>

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

<math>lim_{h\rarr0}\frac{nx^{n-1}h+nx\frac{1}{2}n(n-1)x^{n-2}h^2+ ...}{h}</math>

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>lim_{h\rarr0}(nx^{n-1}+nx\frac{1}{2}n(n-1)x^{n-2}h+ ...)</math>

Right, now we'll just make h (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\frac{1}{2}n(n-1)x^{n-2}*0+ ... </math>

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