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-9e^(7x) over 7x-3 find derivative!
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\[\frac{-9e^{7x}}{7x-3}\]
use quotient rule and chain rule. quotient rule is \[(\frac{f}{g})'=\frac{gf'-fg'}{g^2}\] with \[f(x)=e^{7x},f'(x)=7e^{7x},g(x)=7x-3,g'(x)=7\] and you can multiply by that annoying -9 at the end
you get \[-9\times\frac{(7x-3)\times 7e^{7x}-e^{7x}\times 7}{(7x-3)^2}\]
you didnt finish the other one
what other one?
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the problem that I posted
i can't find it. link
http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e83dec60b8bc11dd551cf85
i already solved it but apparently is wrong
i wrote it out
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