Question

attempt 1: 10 attempts remaining. differentiate $f(t)=\frac{6 + ln t}{6-ln t}$. $f(t)=$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 6+\ln t$, so $u'=\frac{1}{t}$, and $v = 6-\ln t$, so $v'=-\frac{1}{t}$.

Step2: Substitute $u$, $u'$, $v$, $v'$ into quotient - rule

$f'(t)=\frac{\frac{1}{t}(6 - \ln t)-(6+\ln t)(-\frac{1}{t})}{(6 - \ln t)^{2}}$.

Step3: Simplify the numerator

$\frac{1}{t}(6 - \ln t)-(6+\ln t)(-\frac{1}{t})=\frac{6}{t}-\frac{\ln t}{t}+\frac{6}{t}+\frac{\ln t}{t}=\frac{12}{t}$.

Answer:

$\frac{12}{t(6 - \ln t)^{2}}$