# If y=\frac{8}{3x^{2}}, what is the value of y'?

If $y=\frac{8}{3{x}^{2}}$, what is the value of y'?
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Donald Erickson
${y}^{\prime }=\frac{0×3{x}^{2}-6x×8}{{\left(3{x}^{2}\right)}^{2}}$
${y}^{\prime }=\frac{-48x}{9{x}^{4}}$
${y}^{\prime }=-\frac{16}{3{x}^{3}}$
We get the same derivative (since $-\frac{16}{3}{x}^{-3}=-\frac{16}{3{x}^{3}}$), using a different method.
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Judith Couch
$\frac{8}{3{x}^{2}}=\frac{8}{3}\cdot \frac{1}{{x}^{2}}=\frac{8}{3}{x}^{-2}$
${y}^{\prime }=-\frac{16}{3}{x}^{-3}.$
Another possibility (that I find needlessly complicated) is to use the product rule on
$\frac{8}{3}\cdot {x}^{-2}$
${y}^{\prime }=\frac{d}{dx}\left(\frac{8}{3}\right)\cdot {x}^{-2}+\frac{8}{3}\cdot \frac{d}{dx}\left({x}^{-2}\right)$
$=0\cdot {x}^{-2}+\frac{8}{3}\cdot \left(-2x\cdot -3\right)$
$=-\frac{16}{3}{x}^{-3}$