Here's the question you clicked on:
calyne
Differentiate the function F(t) = ln [(2t+1)^3 / (3t-1)^4]
\[\ln \left( {(2t+1)^3 \over (3t -1)^4} \right) = \ln(2t+1)^3 - \ln (3t-1)^4\]
\[u = 2t+1|| v = 3t -1\]\[{du \over dt} = 2||{dv \over dt} = 3\] \[f(t) = \ln u - \ln v\] \[df(t)' = {1 \over u} {du \over dt} - {1\over v}{dv \over dt}\] \[df(t)' = {2 \over 2t+1} - {3 \over 3t-1} = {-5 \over (2t+1)(3t-1)}\]
uh what about the exponents
oh so for logs you don't drop the exponent one right ok
Yeah apologies I forgot to include them!\[f(t) = (\ln u)^3 - (\ln v)^4 = 3\ln u - 4 \ln v\]