Q.Prove that the function $f$ given by $f(x) = \log |\cos x|$ is decreasing on $\left(0, \frac{\pi}{2}\right)$ and increasing on $\left(\frac{3\pi}{2}, 2\pi\right)$.

Concept: Monotonic Function Analysis — we check the sign of $f'(x)$ on each interval.

Step 1: Differentiate $f(x) = \log |\cos x|$. For $x$ in the given intervals, $\cos x$ is positive on $(0, \frac{\pi}{2})$ and positive on $(\frac{3\pi}{2}, 2\pi)$, so $|\cos x| = \cos x$.

Thus

Step 2: On $(0, \frac{\pi}{2})$, $\tan x > 0$, so $f'(x) = -\tan x < 0$. Hence $f$ is decreasing. …

To prove monotonic behaviour of a function, the most direct tool is the sign of its derivative. If $f'(x) > 0$ on an interval, $f$ is strictly increasing there; if $f'(x) < 0$, it is strictly decreasing. The function here is $f(x) = \log |\cos x|$, which involves an absolute value — but since we are only considering intervals where $\cos x$ does not change sign, we can drop the absolute value carefully.

Let’s work through it.

1. Domain and sign of $\cos x$ on the given intervals

   On $\left(0, \frac{\pi}{2}\right)$, $\cos x > 0$, so $|\cos x| = \cos x$.

   On $\left(\frac{3\pi}{2}, 2\pi\right)$, $\cos x > 0$ as well (since cosine is positive in the fourth quadrant). So on both intervals, $f(x) = \log(\cos x)$ — no absolute value complications.

2. Differentiate

   Using the chain rule:

3. Sign of $f'(x)$ on $\left(0, \frac{\pi}{2}\right)$

   On this interval, $\tan x > 0$ (since both $\sin x$ and $\cos x$ are positive). Therefore $f'(x) = -\tan x < 0$.

   A negative derivative means $f$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$.

4. Sign of $f'(x)$ on $\left(\frac{3\pi}{2}, 2\pi\right)$

   Here, $\sin x$ is negative (since sine is negative in the fourth quadrant) and $\cos x$ is positive. So $\tan x = \frac{\sin x}{\cos x} < 0$.

   Hence $f'(x) = -(\text{negative}) = \text{positive}$, i.e. $f'(x) > 0$.

   A positive derivative means $f$ is strictly increasing on $\left(\frac{3\pi}{2}, 2\pi\right)$.

5. Why we don’t need the absolute value inside the log …

Method: Differentiating $\log|g(x)|$ and Handling the Sign of the Inner Function

This method extends the log-of-a-function technique to cases where an absolute value appears inside the logarithm — necessary whenever the inner function (like $\cos x$) is negative somewhere in the domain you're interested in.

Steps

Step 1: On each interval you are asked about, first determine the actual sign of the inner function.

Check whether $\cos x$ (or whichever function is inside) is positive or negative throughout that specific interval — this decides whether $|\cos x|=\cos x$ or $|\cos x|=-\cos x$ there.

Step 2: Differentiate using the chain rule; the derivative of $\log|g(x)|$ simplifies to the same clean formula regardless of the sign of $g(x)$.

This identity holds whether $g(x)$ is positive or negative (as long as it's nonzero), because the absolute value's own derivative contributes a sign that exactly cancels with the sign flip from writing $-g(x)$ instead of $g(x)$.

Step 3: Substitute the specific inner function's derivative and simplify. …

Common Mistakes

Mistake 1: Assuming the absolute value forces a case-split into two different derivative formulas.

Why it's wrong: it seems natural to write $\log(\cos x)$ for one region and $\log(-\cos x)$ for another with two different-looking derivatives, but both cases collapse to the same formula $g'(x)/g(x)$ once differentiated correctly — unnecessary case-splitting wastes time and invites sign errors. Correct approach: differentiate $\log|g(x)|$ directly as $g'(x)/g(x)$, and only use the sign check to interpret the interval, not to derive a different formula.

Mistake 2: Misjudging the sign of $\tan x$ in the fourth quadrant, i.e. on $\left(\tfrac{3\pi}{2},2\pi\right)$. …