Differentiating Functions: Difference between revisions

Revision as of 06:05, 28 October 2005

Revision as of 06:12, 28 October 2005

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

<math>lim_{h\rarr0}\frac{f(x+dx)-f(x)}{dx}</math> the more delta x (dx)h 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>lim_{h\rarr0}\frac{(x+dx)^n-x^n}{dxh}</math>

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

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

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

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

Right, now we'll just make the delta xh (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):

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So, that would basically mean that:

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

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

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

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