By taking the natural log on both sides we can see that: ln(f(x)) = ln(x)^2 This is a more familiar expression that we know how to differentiate LHS: f '(x)/f(x), RHS: 2*ln(x)/x By rearranging this we can see that f '(x) = f(x)2ln(x)/x Substituting our original f(x) expression back into this we find that: f '(x) = x^ln(x)2ln(x)/x = x^(ln(x)-1)2ln(x).
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