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anonymous
 4 years ago
Differentiate the function F(t) = ln [(2t+1)^3 / (3t1)^4]
anonymous
 4 years ago
Differentiate the function F(t) = ln [(2t+1)^3 / (3t1)^4]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\ln \left( {(2t+1)^3 \over (3t 1)^4} \right) = \ln(2t+1)^3  \ln (3t1)^4\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[u = 2t+1 v = 3t 1\]\[{du \over dt} = 2{dv \over dt} = 3\] \[f(t) = \ln u  \ln v\] \[df(t)' = {1 \over u} {du \over dt}  {1\over v}{dv \over dt}\] \[df(t)' = {2 \over 2t+1}  {3 \over 3t1} = {5 \over (2t+1)(3t1)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0uh what about the exponents

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1334352312785:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh so for logs you don't drop the exponent one right ok

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah apologies I forgot to include them!\[f(t) = (\ln u)^3  (\ln v)^4 = 3\ln u  4 \ln v\]
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