such that if

$\displaystyle f(x) = x^{p/q}$ then $\displaystyle f'(x)={p/q}x^{p/q-1}$

using the definition of a derivative (the 'lim as h tends to zero' one),the formula $\displaystyle a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-1}b+...+b^{n-1})$

and fact that the derivative of $\displaystyle f(x)=x^{1/q}$ is $\displaystyle 1/qx^{1/q-1}$ (proved earlier using the above formula, where the "h" falls out nicely)

I have achieved the result implicitly, but did not use the def-n of the derivative or the $\displaystyle a^{n}-b^{n}$ formula