If $f(x)= x^2-x$ and $g(x) = x^( 1/3 )$ , what is $f'(g(x))$ ?

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Answer 1 · 2016-12-21T14:23:21

$1/3x^(-1/3)(2-x^(-1/3))$

$f(g)=g^2-g=(x^(1/3))^2-x^(1/3)=x^(2/3)-x^(1/3)$

$f'(g(x)$

$= (x^(2/3)-x^(1/3))'$

$=2/3x^(-1/3)-1/3x^(-2/3)$

$=1/3x^(-1/3)(2-x^(-1/3))$

If f(x)= x^2-x f(x)= x^2-x and g(x) = x^( 1/3 ) g(x) = x^( 1/3 ) , what is f'(g(x)) f'(g(x)) ?