If $f(x) = |\cos x - \sin x|$, then $f'\left(\dfrac{\pi}{3}\right) = $ __________.

If f(x) = |cos x - sin x|, then f'((π)/(3)) = ____ – Class 12 Mathematics

NCERT Exemplar · Q74

f(x) = |\cos x - \sin x|

Kerala DhseShort· 1mImportance★★★★★

Appeared in past exams:KEAM 2024· Set eng-2024-0609· 4mexact

90% · 254/281 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Differentiability of Absolute Value

Differentiability of the Absolute Value Function

Start with something familiar: the absolute value of $x$ , written $|x|$ , is its distance from zero on the number line. So $|3| = 3$ , $|-5| = 5$ , and $|0| = 0$ . Graphically, it looks like a V-shape — two straight lines meeting at the origin.

Differentiability is about whether a function has a well-defined slope (derivative) at a point. For smooth curves like $x^2$ or $\sin x$ , the slope exists everywhere. But the absolute value function has a sharp corner at $x = 0$ — and that corner is the whole story.

Intuition: Why the corner matters

Walk along $y = |x|$ from left to right. Approaching $x = 0$ from the left, the slope is $-1$ (the line goes downward). Leaving $x = 0$ to the right, the slope is suddenly $+1$ (the line goes upward). At $x = 0$ , there's no single slope — it changes abruptly. That's why $|x|$ is not differentiable at $x = 0$ . Everywhere else — for $x < 0$ and $x > 0$ — the graph is a straight line with constant slope, so $|x|$ is differentiable at every point except $x = 0$ .

Note

A function must be continuous to be differentiable, but continuity alone isn't enough. The absolute value function is continuous at $x = 0$ (no break), yet fails to be differentiable there because of the sharp corner.

The precise statement

Let $f(x) = |x|$ . Then:

For $x > 0$ : $f(x) = x$ , so $f'(x) = 1$ .
For $x < 0$ : $f(x) = -x$ , so $f'(x) = -1$ .
At $x = 0$ : the derivative does not exist, because the left-hand and right-hand derivatives are different numbers.

f'(0) = \lim_{h \to 0} \frac{|0+h| - |0|}{h} = \lim_{h \to 0} \frac{|h|}{h}

Watch out

A common mistake is to think that because $|x|$ is continuous at $x=0$ , it must be differentiable there. Continuity is necessary for differentiability, but not sufficient. The absolute value function is the classic counterexample.

The bigger picture …

Start your 14-day free trial to unlock the full solution →

At $x=\dfrac{\pi}{3}$ , $\cos x-\sin x<0$ , so $f(x)=\sin x-\cos x$ locally; hence $f'\left(\dfrac{\pi}{3}\right)=\dfrac{1+\sqrt{3}}{2}$ .

Fix the branch. At $x=\dfrac{\pi}{3}$ , $\cos\dfrac{\pi}{3}=\dfrac12$ and $\sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}$ , so

\cos x-\sin x=\frac{1-\sqrt3}{2}<0.

Near this point $|\cos x-\sin x|=-(\cos x-\sin x)=\sin x-\cos x$ .

Differentiate that branch.

$f'(x)=\cos x+\sin x.$ …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2026Set eng-2026-04194 marksMCQ
$f(x) = x|x|$
›Reveal solutionSolution
$t(x)=x+[x]$ has a jump discontinuity at $x=0$ from the greatest-integer part, so it cannot be differentiable there. Statement (E) is untrue.

Check the claims:

(A) $f(x)=x|x|$ has $f'(x)=2|x|$ , differentiable on $(-1,1)$ including $0$ . True.

(B) $\sqrt{|x|}=\sqrt{x}$ on $(4,5)$ (positive $x$ ), smooth. True.

(C) On $(2,3)$ both $|x-2|,|x+3|$ are linear — differentiable. True. …
Start your 14-day free trial to see the full solution →
KEAM 2026Set eng-2026-04194 marksMCQ
$y = \frac{3x^3 - 2x^2 + x}{|x|}, x \neq 0$
›Reveal solutionSolution

Near $x=-2$ (negative), $|x|=-x$ , giving $y=-3x^2+2x-1$ ; its derivative at $-2$ is $14$ .

Simplify for $x<0$ . With $|x|=-x$ ,

$y=\frac{3x^3-2x^2+x}{-x}=-(3x^2-2x+1)=-3x^2+2x-1.$ …
Start your 14-day free trial to see the full solution →
KEAM 2026Set eng-2026-04224 marksMCQ
$f(x) = \frac{|x|}{x^2}$
›Reveal solutionSolution

Simplify $f$ near $x=2$ (where $x>0$ ) before differentiating.

For $x>0$ , $|x|=x$ , so $f(x)=\frac{x}{x^2}=\frac1x=x^{-1}$ . …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04234 marksMCQ
$f(x) = |x^2 + x - 6|$
›Reveal solutionSolution
The corners of |x^2+x-6| sit at its roots x=-3 and x=2, so a^2+b^2 = 9+4 = 13.

Concept and Intuition

An absolute value |g(x)| develops a sharp corner (non-differentiability) exactly where g(x) changes sign, i.e. at the simple real roots of g. There the left and right derivatives are equal in magnitude but opposite in sign, so f'(x) does not exist.

Step-by-Step Solution

Factor the inside: x^2 + x - 6 = (x+3)(x-2).

The quadratic changes sign at its simple roots x = -3 and x = 2, so f(x) = |x^2+x-6| has corners there.

Thus a = -3, b = 2 (in either order). …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04264 marksMCQ
$f(x) = 10 - |x-5|, x \in \mathbb{R}$
›Reveal solutionSolution

$f(x)=10-|x-5|$ has a sharp corner where $x-5=0$ , i.e. at $x=5$ .

The absolute-value function $|x-5|$ is differentiable everywhere except at $x=5$ , where it has a corner (left slope $-1$ , right slope $+1$ ). Subtracting it from the constant $10$ preserves that no …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04274 marksMCQ
$f(x) = |x^2 - 3x + 2|$
›Reveal solutionSolution

$f(x)=|x^2-3x+2|$ fails to be differentiable only at the roots of the quadratic inside the modulus, namely $x=1$ and $x=2$ .

The quadratic factors as $x^2-3x+2=(x-1)(x-2)$ , so $f(x)=|(x-1)(x-2)|$ .

A modulus function $|g(x)|$ is differentiable everywhere $g$ is differentiable except at simple zeros of $g$ , where the graph is reflected upward and forms a sharp corner. Here $g(x)=(x-1)(x-2)$ is a polynomial (differentiable everywhere) with simple roots at $x=1$ and $x=2$ . …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04274 marksMCQ
$f(x) = x|x|$
›Reveal solutionSolution

On $x<0$ , $f(x)=x|x|=-x^2$ , giving $f'(x)=-2x$ , so $f'(-10)=20$ .

Write $f(x)=x|x|$ piecewise. For $x<0$ we have $|x|=-x$ , so

$f(x)=x(-x)=-x^2.$ …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04294 marksMCQ
$f(x) = |\cos x - \sin x|$
›Reveal solutionSolution

Near $x=\frac{\pi}{6}$ , $\cos x-\sin x>0$ , so $f(x)=\cos x-\sin x$ and $f'(x)=-\sin x-\cos x$ , giving $-\frac{\sqrt3+1}{2}$ .

At $x=\frac{\pi}{6}$ : $\cos\frac{\pi}{6}=\frac{\sqrt3}{2}$ , $\sin\frac{\pi}{6}=\frac12$ , so $\cos x-\sin x=\frac{\sqrt3-1}{2}>0$ . Thus locally $|\cos x-\sin x|=\cos x-\sin x$ , and

$f'(x)=-\sin x-\cos x.$ …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04294 marksMCQ
$f(x) = |x^2 - 1|$
›Reveal solutionSolution

Near $x=\frac32$ , $x^2-1>0$ , so $f(x)=x^2-1$ and $f'(x)=2x=3$ .

At $x=\frac32$ , $x^2-1=\frac94-1=\frac54>0$ , so locally $|x^2-1|=x^2-1$ . Then …
Start your 14-day free trial to see the full solution →
KEAM 2024Set eng-2024-06084 marksMCQ
$f(x)=x|x|$
›Reveal solutionSolution
$f(x)=x|x|$ equals $x^2$ for $x\ge0$ and $-x^2$ for $x<0$ ; its derivative is $2|x|$ , so $f'(-1)+f'(1)=2+2=4$ .

Write the function piecewise:
$f(x)=\begin{cases}x^2,& x\ge 0\\ -x^2,& x<0.\end{cases}$
Differentiate each branch:

For $x>0$ : $f'(x)=2x$ , so $f'(1)=2$ . …
Start your 14-day free trial to see the full solution →
KEAM 2024Set eng-2024-06094 marksMCQ
$f(x) = |\cos x - \sin x|$
›Reveal solutionSolution

On $(\frac{\pi}{4},\frac{\pi}{2})$ we have $\sin x>\cos x$ , so $f(x)=\sin x-\cos x$ and $f'(x)=\cos x+\sin x$ ; evaluating at $\frac{\pi}{3}$ gives $\frac{\sqrt3+1}{2}$ .

For $x\in\left(\frac{\pi}{4},\frac{\pi}{2}\right)$ , $\sin x>\cos x$ , so $\cos x-\sin x<0$ and
$f(x)=|\cos x-\sin x|=\sin x-\cos x.$
Then $f'(x)=\cos x+\sin x$ . …
Start your 14-day free trial to see the full solution →

←Previous questionQ73 — If f(x) = |cos x|, then f'((π)/(4)) = _____.→Next questionQ75 — For the curve √(x) + √(y) = 1, (dy)/(dx) at ((1)/(4), (1)/(4)) is ____…

Q.If $f(x) = |\cos x - \sin x|$ , then $f'\left(\dfrac{\pi}{3}\right) =$ __________.

Concept understanding — Differentiability of Absolute Value

Differentiability of the Absolute Value Function

Intuition: Why the corner matters

The precise statement

The bigger picture …

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

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Q.If f(x)=∣cos⁡x−sin⁡x∣f(x) = |\cos x - \sin x|f(x)=∣cosx−sinx∣, then f′(π3)=f'\left(\dfrac{\pi}{3}\right) = f′(3π​)= __________.

🎓Unlock everything free for 14 days

Q.If $f(x) = |\cos x - \sin x|$ , then $f'\left(\dfrac{\pi}{3}\right) =$ __________.

Unlock everything free for 14 days