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

1. anonymous

$\ln \left( {(2t+1)^3 \over (3t -1)^4} \right) = \ln(2t+1)^3 - \ln (3t-1)^4$

2. anonymous

$u = 2t+1|| v = 3t -1$${du \over dt} = 2||{dv \over dt} = 3$ $f(t) = \ln u - \ln v$ $df(t)' = {1 \over u} {du \over dt} - {1\over v}{dv \over dt}$ $df(t)' = {2 \over 2t+1} - {3 \over 3t-1} = {-5 \over (2t+1)(3t-1)}$

3. anonymous

4. anonymous

|dw:1334352312785:dw|

5. anonymous

oh so for logs you don't drop the exponent one right ok

6. anonymous

Yeah apologies I forgot to include them!$f(t) = (\ln u)^3 - (\ln v)^4 = 3\ln u - 4 \ln v$

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