Difficult Calculus problem: Solve this problem via binomial expansion theorem?

Maybe he wanted you to use the binomial expansion theorem to calculate the derivative, using the def'n of a derivative:

With h small, (x + h)^(p/q) ≈ x^(p/q) [1 + (p/q)(h/x) + O(h/x)^2 +...

Thus: limit h->0 [f(x + h) - f(x)]/h ≈ [x^(p/q) + (p/q)hx^(p/q - 1) +O(h/x)^2 - x^(p/q)] / h

≈ (p/q)x^((p/q)-1) +O(h)

Thus if f(x) = x^(p/q), f ' (x) = (p/q) x^((p/q)-1)

QED

I don't know what your application is, but the expansion can be found by taking h^(p/q) out as a factor.

(x+h)^(p/q)=h^(p/q)*(1+x/h)^(p/q)

=h^(p/q)*(1+px/hq+(p/q)(-1+p/q)(x^2)/(2h^2)+....)

Using calculus you can get this by differentiating y(x)=(x+h)^(p/q) repeatedly wrt x, and saying well if this function were a polynomial the coefficients of the first n terms (increasing powers of x) would be: y(0), y'(0), y''(0)/2,...

...,y[n](0)/n!.respectively.

where y[n]=y differentiated n times . This is the idea behind taylor series as I understand it.

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You need to be more explicit in what you want; I can't guess what it is

I'm not sure what the question is

do you mean 'how do you expand' this expression?