daveed said:
okay... could someone show me how to find the derivative of 5/(x^2/5) using ONLY the definition of a derivative( (f(x+h)-f(x))/h as h approaches 0), and not using any of the rules or proving the rules
strictly algebraicly...(probably spelled that wrong)
After about an hour (I was about to just give up) of messing around I had an idea. I needed to know if I could factor x^5 - y^5 and it turns out that I can!
For your reference
x^5-y^5 = (x-y)(x^4 + x^3y + x^2y^2 +xy^3 + y^4)
what you have is
<br />
5\lim_{h\rightarrow 0} \frac {x^\frac {2} {5} - (x+h)^\frac {2} {5}} {h(x^2 + 2hx)^\frac {2} {5}}
Now the trick to getting rid of that h in the denominator is to figure out how you could go about adding the two parts of the numerator.
Start by noticing that the numerator is a difference of squares
<br />
x^\frac {2} {5} - (x+h)^\frac {2} {5}<br />
Which of course you can factor out to
<br />
(x^\frac {1} {5} + (x+h)^\frac {1} {5})(x^\frac {1} {5} - (x+h)^\frac {1} {5})<br />
Now all you have got to do is figure out how to get ride of that nasty little h in the denominator. That is not so easy unless you multiply the entire quotient by 1. That is, multiply by
<br />
\frac {x^\frac {4} {5} + x^\frac {3} {5}(x+h)^\frac {1} {5} + x^\frac {2} {5}(x+h)^\frac {2} {5} + x^\frac {1} {5}(x+h)^\frac {3} {5} + (x+h)^\frac{4} {5}} {x^\frac {4} {5} + x^\frac {3} {5}(x+h)^\frac {1} {5} + x^\frac {2} {5}(x+h)^\frac {2} {5} + x^\frac {1} {5}(x+h)^\frac {3} {5} + (x+h)^\frac{4} {5}}<br />
Which, if you will recall from our earlier results, we can easily multiply the (x^\frac {1} {5} - (x+h)^\frac {1} {5}) with the x^\frac {4} {5} + x^\frac {3} {5}(x+h)^\frac {1} {5} + x^\frac {2} {5}(x+h)^\frac {2} {5} + x^\frac {1} {5}(x+h)^\frac {3} {5} + (x+h)^\frac{4} {5} and that will get rid of the fifth root and thus will give us a nice and neat numerator from which you can factor out the h. If you have gotten this far the rest is cake.
Regards
