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

The key idea is that absolute value functions create V-shaped graphs with a corner at the point where the expression inside becomes zero. Differentiability fails at such corners because the left-hand and right-hand slopes differ. Here we have two absolute values, so we expect trouble at both $x=0$ and $x=1$.

Let’s work through it step by step.

1. Rewrite the function piecewise

   The absolute value $|x|$ changes behavior at $x=0$, and $|x-1|$ changes at $x=1$. So the whole function has three natural intervals:

   • For $x < 0$: $|x| = -x$ and $|x-1| = -(x-1) = 1-x$. So $f(x) = -x + (1-x) = 1 - 2x$.
   • For $0 \le x < 1$: $|x| = x$ and $|x-1| = 1-x$. So $f(x) = x + (1-x) = 1$.
   • For $x \ge 1$: $|x| = x$ and $|x-1| = x-1$. So $f(x) = x + (x-1) = 2x - 1$.

   Thus:

2. Check differentiability at $x=0$

   Differentiability requires that the left-hand derivative equals the right-hand derivative.

   • Left-hand derivative at 0: For $x<0$, $f'(x) = -2$. So the slope from the left is $-2$.
   • Right-hand derivative at 0: For $0 < x < 1$, $f(x)=1$, so $f'(x)=0$. The slope from the right is $0$. Since $-2 \neq 0$, the derivatives do not match. Hence $f$ is not differentiable at $x=0$.

3. Check differentiability at $x=1$

   • Left-hand derivative at 1: For $0 \le x < 1$, $f(x)=1$, so $f'(x)=0$.
   • Right-hand derivative at 1: For $x>1$, $f'(x)=2$. …

1. Structure of the function

Think of $y=\big||x|-1\big|$ in two layers: first $|x|$ (a corner at $x=0$), then subtract 1 and take the absolute value again (a corner wherever $|x|-1=0$, i.e. $x=\pm1$). So the candidate non-differentiable points are $x=-1,0,1$.

2. Check each point

• At $x=0$: near $0$, $|x|-1<0$, so $y = 1-|x|$. Left derivative ($h\to0^-$): $\dfrac{d}{dx}(1+x) = 1$. Right derivative ($h\to0^+$): $\dfrac{d}{dx}(1-x) = -1$. They differ → not differentiable.

• At $x=1$: for $x$ slightly less than 1, $|x|-1<0$ so $y=1-x$, slope $-1$. For $x$ slightly more than 1, $|x|-1>0$ so $y=x-1$, slope $+1$. They differ → not differentiable. …

The only non-differentiable point of $e^{-|x-1|}$ is the corner at $x=1$. But the domain here is the open interval $(1,\infty)$, which excludes $x=1$. For every $x>1$:

$f(x)=e^{-(x-1)},\qquad f'(x)=-e^{-(x-1)},$ …