Q.If $f(x) = |\cos x|$, then $f'\left(\dfrac{\pi}{4}\right) =$ __________.

The key idea is that $| \cdot |$ is differentiable except where its argument is zero. Here, $\cos x$ is positive near $x = \frac{\pi}{4}$, so the absolute value can be dropped locally.

Step 1: For $x$ near $\frac{\pi}{4}$, $\cos x > 0$, so $f(x) = \cos x$.

Step 2: Then $f'(x) = -\sin x$ in that neighbourhood. …

The key to differentiating an absolute value function like $f(x) = |\cos x|$ is understanding where the expression inside the absolute value is positive, negative, or zero. The absolute value function $|u|$ has derivative $u'$ when $u > 0$, derivative $-u'$ when $u < 0$, and is not differentiable when $u = 0$ (unless $u'$ is also zero, which is a special case).

Here, $u = \cos x$. At $x = \pi/4$, we have $\cos(\pi/4) = \frac{1}{\sqrt{2}} > 0$. So near $x = \pi/4$, the absolute value does nothing — $|\cos x| = \cos x$ locally. That means the derivative at that point is simply the derivative of $\cos x$.

Let’s work through it step by step.

1. Check the sign of $\cos x$ at $x = \pi/4$.

   $\cos(\pi/4) = \frac{\sqrt{2}}{2} > 0$. Since $\cos x$ is continuous, it remains positive in a small interval around $\pi/4$. Therefore, in that neighbourhood, $f(x) = |\cos x| = \cos x$.

2. Differentiate the simplified function.

   For $x$ near $\pi/4$, $f(x) = \cos x$, so $f'(x) = -\sin x$.

3. Evaluate at $x = \pi/4$.

   $f'(\pi/4) = -\sin(\pi/4) = -\frac{\sqrt{2}}{2}$. …

Method: Differentiating an Absolute Value Function at a Specific Point (Sign-Check Method)

This method solves "find $f'(a)$ where $f(x)=|g(x)|$" problems by removing the absolute value locally, using the sign of $g$ at the given point.

Steps

Step 1: Evaluate the inside expression at the given point

Compute $g(a)$, the quantity inside the modulus, at the point where the derivative is required.

Step 2: Determine its sign

If $g(a)>0$, then by continuity of $g$, the expression stays positive in a small neighbourhood of $a$, so $|g(x)|=g(x)$ locally — the modulus does nothing there. If $g(a)<0$, then $g$ stays negative nearby, so $|g(x)|=-g(x)$ locally.

Step 3: Differentiate the branch that applies …

Common Mistakes

Mistake 1: Differentiating without checking the sign of $\cos x$ first

Students often jump straight to $f'(x)=-\sin x$ (or, worse, $\sin x$) without confirming whether $\cos x$ is positive or negative near $x=\frac{\pi}{4}$. Why it's wrong: $|\cos x|$ equals $\cos x$ only where $\cos x\ge 0$, and equals $-\cos x$ where $\cos x<0$ — using the wrong branch flips the sign of the final answer. Correct approach: evaluate $\cos\frac{\pi}{4}=\frac{\sqrt2}{2}>0$ first, confirming the absolute value can be dropped locally, THEN differentiate.

Mistake 2: Misapplying the general $|u|$-derivative formula …