How do you find the derivative of $f(x)=x^3-3x$ ?

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Answer 1 · 2016-02-12T14:27:49

$3x^2 - 3$

using

If $y = ax^n then dy/dx = nax^(n-1)$

here $f(x) = x^3 - 3x$

hence f'(x) = $3x^(3-1) - 3x^(1-1) = 3x^2 -3x^0 = 3x^2 - 3$

(since $x^0 = 1)$

How do you find the derivative of f(x)=x^3-3xf(x)=x^3-3x?