5. given that $s(t)=sqrt{\frac{1 + sin t}{1+cos t}}$ show that $s^{prime}(t)=\frac{cos t+sin t +…

Explanation:

Step1: Rewrite the function

Rewrite (s(t)=\sqrt{\frac{1 + \sin t}{1+\cos t}}=\left(\frac{1+\sin t}{1 + \cos t}\right)^{\frac{1}{2}})

Step2: Apply the chain - rule and quotient - rule

The chain - rule: If (y = u^{\frac{1}{2}}) and (u=\frac{1+\sin t}{1+\cos t}), then (y^\prime=\frac{1}{2}u^{-\frac{1}{2}}\cdot u^\prime) The quotient - rule: If (u=\frac{f(t)}{g(t)}) where (f(t)=1+\sin t), (f^\prime(t)=\cos t) and (g(t)=1+\cos t), (g^\prime(t)=-\sin t), then (u^\prime=\frac{f^\prime(t)g(t)-f(t)g^\prime(t)}{g^{2}(t)}) [ \begin{align*} u^\prime&=\frac{\cos t(1 + \cos t)-(1+\sin t)(-\sin t)}{(1+\cos t)^{2}}\ &=\frac{\cos t+\cos^{2}t+\sin t+\sin^{2}t}{(1+\cos t)^{2}}\ \end{align*} ] Since (\sin^{2}t+\cos^{2}t = 1), then (u^\prime=\frac{\cos t+\sin t + 1}{(1+\cos t)^{2}})

Step3: Find (s^\prime(t))

[ \begin{align*} s^\prime(t)&=\frac{1}{2}\left(\frac{1+\sin t}{1+\cos t}\right)^{-\frac{1}{2}}\cdot\frac{\cos t+\sin t + 1}{(1+\cos t)^{2}}\ &=\frac{\cos t+\sin t + 1}{2\sqrt{\frac{1+\cos t}{1+\sin t}}\cdot(1+\cos t)^{2}}\ &=\frac{\cos t+\sin t + 1}{2\sqrt{1+\sin t}(1+\cos t)^{\frac{3}{2}}} \end{align*} ]

Answer:

We have shown that (s^\prime(t)=\frac{\cos t+\sin t + 1}{2\sqrt{1+\sin t}(1+\cos t)^{\frac{3}{2}}}) by using the chain - rule and quotient - rule for differentiation.