Let f(x) be a differentiable function such that xf(x)+f(x^2)=2 for all x>0 find f'(1)?
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Differentiate: xf'(x)+f(x)+2xf'(x^2)=0; f(x)+f(x^2)/x=2/x, so xf'(x)+2/x-f(x^2)/x+2xf'(x^2)=0.

Putting x=1, we have f'(1)+2-f(1)+2f'(1)=0. To find f(1), substitute x=1 in xf(x)+f(x^2)=2: f(1)+f(1)=2, so f(1)=1. Substitute for f(1) in the derivative: 3f'(1)+2-1=0, so f'(1)=-1/3.

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