differentiate $a(t)=lnleft(left(\frac{1 - sin t}{1+sin t}\right)^{3}\right)$. it may be to your advantage to…

Explanation:

Step1: Use logarithm property

Using the property $\ln(a^b)=b\ln(a)$, we can rewrite $a(t)$ as $a(t) = 3\ln\left(\frac{1 - \sin t}{1+\sin t}\right)$.

Step2: Use the property $\ln(\frac{u}{v})=\ln(u)-\ln(v)$

$a(t)=3[\ln(1 - \sin t)-\ln(1 + \sin t)]$.

Step3: Differentiate term - by - term

The derivative of $\ln(u)$ with respect to $t$ is $\frac{u'}{u}$ by the chain - rule. For $y_1 = \ln(1 - \sin t)$, let $u = 1-\sin t$, then $u'=-\cos t$, and $\frac{d}{dt}\ln(1 - \sin t)=\frac{-\cos t}{1 - \sin t}$. For $y_2=\ln(1 + \sin t)$, let $u = 1+\sin t$, then $u'=\cos t$, and $\frac{d}{dt}\ln(1 + \sin t)=\frac{\cos t}{1 + \sin t}$.

Step4: Calculate $a'(t)$

$a'(t)=3\left[\frac{-\cos t}{1 - \sin t}-\frac{\cos t}{1 + \sin t}\right]$. Find a common denominator $(1 - \sin t)(1 + \sin t)=1-\sin^{2}t=\cos^{2}t$. $a'(t)=3\left[\frac{-\cos t(1 + \sin t)-\cos t(1 - \sin t)}{\cos^{2}t}\right]$. Expand the numerator: $-\cos t(1 + \sin t)-\cos t(1 - \sin t)=-\cos t-\cos t\sin t-\cos t+\cos t\sin t=-2\cos t$. So $a'(t)=3\times\frac{-2\cos t}{\cos^{2}t}=-\frac{6}{\cos t}=-6\sec t$.

Answer:

$-6\sec t$