Find y' for: (a) y = ln tan x/2; (b) y = cos x ·ln tan x

Find $y'$ for: (a) $y = \ln\tan\dfrac{x}{2}$; (b) $y = \cos x\cdot\ln\tan x$.

Step 1: Part (a): apply the chain rule

With $u = \tan\dfrac{x}{2}$, the inner derivative carries both the tangent rule and the factor $\tfrac12$: \[u' = \frac{1}{\cos^{2}\frac{x}{2}}\cdot\frac{1}{2}, \qquad y' = \frac{u'}{u} = \frac{1}{\tan\frac{x}{2}}\cdot\frac{1}{2\cos^{2}\frac{x}{2}}.\]

Step 2: Part (a): simplify

Writing $\dfrac{1}{\tan\frac{x}{2}} = \dfrac{\cos\frac{x}{2}}{\sin\frac{x}{2}}$, one factor of $\cos\frac{x}{2}$ cancels and the double-angle identity finishes the job: \[y' = \frac{\cos\frac{x}{2}}{2\sin\frac{x}{2}\cos^{2}\frac{x}{2}} = \frac{1}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \frac{1}{\sin x}.\]

\[\text{(a)}\qquad y' = \dfrac{1}{\sin x}\]

Step 3: Part (b): the logarithmic factor

The same computation without the half-angle gives $\bigl(\ln\tan x\bigr)' = \dfrac{1}{\sin x\cos x} = \dfrac{2}{\sin 2x}$.

Step 4: Part (b): apply the product rule

With $u = \cos x$, $u' = -\sin x$, $v = \ln\tan x$, $v' = \dfrac{1}{\sin x\cos x}$, the factor $\cos x$ cancels in the second term: $y' = -\sin x\,\ln\tan x + \cos x\cdot\frac{1}{\sin x\cos x} = -\sin x\,\ln\tan x + \frac{1}{\sin x}$.

\[\text{(b)}\qquad y' = -\sin x\,\ln\tan x + \dfrac{1}{\sin x}\]

Step 5: Verification

\[\text{(a)}\ x = \frac{\pi}{2}:\quad y = \ln\tan\frac{\pi}{4} = 0,\quad y' = \frac{1}{\sin\frac{\pi}{2}} = 1 \;\;\checkmark\]

\[\text{(b)}\ x = \frac{\pi}{4}:\quad \ln\tan\frac{\pi}{4} = 0 \;\Longrightarrow\; y' = 0 + \frac{1}{\sin\frac{\pi}{4}} = \sqrt2 \approx 1.41421 \;\;\checkmark\]