How do you find the derivative for $2 x^{4} - 1$ $2x^4-1$ ?

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Answer 1 · 2015-06-02T07:36:25

$f (x) = 2 x^{4} - 1 = g (x) + h (x)$ $f(x) = 2x^4 - 1 = g(x) + h(x)$ , where $g (x) = 2 x^{4}$ $g(x) = 2x^4$ and $h (x) = - 1$ $h(x) = -1$

$f'(x) = g'(x) + h'(x) = (2x^4)' + (-1)' = 4*2x^(4-1) + 0 = 8x^3$ .

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