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3x^4y^2 / xy^2
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3x^3
3x^3
i'd look it as \[(3x^4y^2)*(x^-1y^-2)\]
then al you have to do is add
1/x is the same as x^-1
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yes - good thinking
I do it that way, i can't stand division... ever since i hit the quotient rule in calc 1
lol - quotient rule is irritating
how do u guys get this answer, im only 14 btw
you don't need quotient rule
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\[f(x)=\frac{h(x)}{g(x)} =>f(x)=h(x) \cdot [g(x)]^{-1}, g(x) \neq 0\]
use product rule instead :)
+chain rule
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