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What is the first derivative of $x^{2} + (\frac{16}{x})$ $x^2 + (16/x)$ ?

1 Answer

May 13, 2018

$y'=2x-16/x^2$

Explanation:

show below

$y=x^2 + (16/x)$

$y'=2x-16/x^2$

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