Find y' for y = ln cos x
Find $y'$ for $y = \ln\cos x$.
Explanation of Method
The companion to Problem 119, with the sign reversed.
Step 1: Apply the chain rule
With $u = \cos x$ and $u' = -\sin x$: $y' = \frac{-\sin x}{\cos x} = -\tan x$.
Step 2: Verification
\[x = 0:\quad y = \ln 1 = 0,\quad y' = -\tan 0 = 0, \text{ the maximum of } y \;\;\checkmark\]
\[x = \frac{\pi}{4}:\quad y' = -1; \qquad \frac{y(\pi/4+10^{-5})-y(\pi/4-10^{-5})}{2\times 10^{-5}} \approx -1.0 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = -\tan x, \qquad -\frac{\pi}{2}+2k\pi < x < \frac{\pi}{2}+2k\pi\,}\]
Remark
Remark. Read backwards, this is the standard integral $\displaystyle\int\tan x\,dx = -\ln\cos x + C$. Note the parallel with Problem 119: the logarithmic derivative of a function is $\dfrac{f'}{f}$, so $\ln\sin x$ gives $\cot x$ and $\ln\cos x$ gives $-\tan x$ — the two cotangent-tangent pairs with opposite signs.