Find y' for y = ln tan x - (1+ cot x)/x

Find $y'$ for $y = \ln\tan x - \dfrac{1+\cot x}{x}$.

Step 1: Differentiate the logarithmic term

As in Problem 131, $\bigl(\ln\tan x\bigr)' = \dfrac{1}{\tan x}\cdot\dfrac{1}{\cos^{2}x}$, which simplifies to

\[ \frac{d}{dx}\ln\tan x = \frac{1}{\sin x\cos x} = \frac{2}{\sin 2x}. \]

Step 2: Differentiate the quotient

With $u = 1+\cot x$, $u' = -\dfrac{1}{\sin^{2}x}$, $v = x$, $v' = 1$: \[\frac{d}{dx}\!\left(\frac{1+\cot x}{x}\right) = \frac{-\dfrac{x}{\sin^{2}x} - \bigl(1+\cot x\bigr)}{x^{2}}.\]

Step 3: Combine

Subtracting that whole expression flips both of its signs: \[y' = \frac{2}{\sin 2x} + \frac{x}{x^{2}\sin^{2}x} + \frac{1+\cot x}{x^{2}} = \frac{2}{\sin 2x} + \frac{1}{x\sin^{2}x} + \frac{1+\cot x}{x^{2}}.\]

Step 4: Verification

\[x = \frac{\pi}{4}:\quad \sin 2x = 1,\ \sin^{2}x = \tfrac12,\ \cot x = 1\]

\[y' = 2 + \frac{2}{\pi/4} + \frac{2}{(\pi/4)^{2}} = 2 + \frac{8}{\pi} + \frac{32}{\pi^{2}} \approx 2 + 2.54648 + 3.24228 = 7.78876\]

\[\frac{y(\pi/4+10^{-5})-y(\pi/4-10^{-5})}{2\times 10^{-5}} \approx 7.78876 \;\;\checkmark\]