#### Answer

$$ y'=15 x^4-14x.$$

#### Work Step by Step

Recall that $(x^n)'=nx^{n-1}$
Since $ y=3x^5-7x^2+4$, then the derivative $ y'$ is given by
$$ y'=3\times5x^{5-1}-2\times7x^{2-1}+0$$
$$=15 x^4-14x$$

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Published by
W. H. Freeman

ISBN 10:
1464125260

ISBN 13:
978-1-46412-526-3

$$ y'=15 x^4-14x.$$

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