Actually, it's perfectly doable from first principles, as long as you can use the general form of the binomial theorem.
I don't think there's any harm here in showing full working, it should be illustrative.
<br />
\begin{array}{l}<br />
f(x) = x - 5(x - 1)^{2/5} \\ <br />
\\ <br />
f'(x) =\lim_{h\rarrow 0} \frac{{f(x + h) - f(x)}}{h} = \frac{{(x + h) - 5((x + h) - 1)^{2/5} - \left( {x - 5(x - 1)^{2/5} } \right)}}{h} \\ <br />
\end{array}<br />
So (I'm going to forego the limit notation here, and just take the limit at the end). f(x+h) = (x + h) - 5(x-1 +h)^{\frac{2}{5}}
f(x+h) = (x + h) - 5(x-1)^{\frac{2}{5}}(1 +\frac{h}{x-1})^\frac{2}{5}
Expand that with Binomial Theorem to a first order approximation (the higher order terms will vanish once the limit is taken anyway). I am including the second order term for clarity.
f(x+h) = (x + h) - 5(x-1)^{\frac{2}{5}}(1 +(\frac{2}{5})\frac{h}{x-1} - (\frac{3}{25})(\frac{h^2}{(x-1)^2} + ...)
Now subtract f(x) from that.
f(x+h) - f(x) = (x + h - x) - 5(x-1)^{\frac{2}{5}}(1 +(\frac{2}{5})\frac{h}{x-1} - 1) - (\frac{3}{25})(\frac{h^2}{(x-1)^2} + ...)
f(x+h) - f(x) = h - 5(x-1)^{\frac{2}{5}}((\frac{2}{5})\frac{h}{x-1} ) - (\frac{3}{25})(\frac{h^2}{(x-1)^2} + ...)
Divide that by h :
\frac{f(x+h) - f(x)}{h} = 1 - 5(x-1)^{(\frac{2}{5})}(\frac{2}{5}\frac{1}{x-1} ) - (\frac{3}{25})(\frac{h}{(x-1)^2} + ...)
Take the limit as h tends to zero, all higher order terms vanish, and you're left with :
f'(x) = 1 - 5(x-1)^{\frac{2}{5}}((\frac{2}{5})\frac{1}{x-1}) = 1 - 2(x-1)^\frac{-3}{5}
I rushed thru this, there may be a bracketing error or a sign error here and there, but I think you should get the general idea.
I don't quite think it's fair to use infinite series before you are even allowed to use the derivative rules! And it isn't something a beginning calc student would be expected to know anyways.
