Introduction
A real-valued function f(x) is differentiable at a point x = c if and only if both its left-hand derivative (LHD) and right-hand derivative (RHD) exist and are equal at that point. For the function f(x) = |cos x| + |sin x|, the modulus operator introduces potential points of non-differentiability where the underlying trigonometric expressions vanish.
Value of the Function at x = π/2
Substituting x = π/2 into f(x):
f(π/2) = |cos(π/2)| + |sin(π/2)| = |0| + |1| = 1.
Evaluation of the Left-Hand Derivative (LHD)
For x approaching (π/2)⁻ from the left, let x = π/2 - h where h > 0 and h → 0:
- Behaviour of cos x: cos(π/2 - h) = sin h > 0, which implies |cos x| = sin h.
- Behaviour of sin x: sin(π/2 - h) = cos h > 0, which implies |sin x| = cos h.
Thus, f(π/2 - h) = sin h + cos h.
By the definition of the left-hand derivative:
f'(π/2⁻) = lim_{h → 0⁺} [f(π/2 - h) - f(π/2)] / (-h)
f'(π/2⁻) = lim_{h → 0⁺} [sin h + cos h - 1] / (-h) = lim_{h → 0⁺} [ - (sin h)/h - (cos h - 1)/h ]
Using standard limits lim_{h → 0} (sin h)/h = 1 and lim_{h → 0} (1 - cos h)/h = 0:
f'(π/2⁻) = -1 - 0 = -1.
Evaluation of the Right-Hand Derivative (RHD)
For x approaching (π/2)⁺ from the right, let x = π/2 + h where h > 0 and h → 0:
- Behaviour of cos x: cos(π/2 + h) = -sin h < 0, which implies |cos x| = -(-sin h) = sin h.
- Behaviour of sin x: sin(π/2 + h) = cos h > 0, which implies |sin x| = cos h.
Thus, f(π/2 + h) = sin h + cos h.
By the definition of the right-hand derivative:
f'(π/2⁺) = lim_{h → 0⁺} [f(π/2 + h) - f(π/2)] / h
f'(π/2⁺) = lim_{h → 0⁺} [sin h + cos h - 1] / h = lim_{h → 0⁺} [ (sin h)/h + (cos h - 1)/h ]
f'(π/2⁺) = 1 + 0 = 1.
Comparison and Conclusion
Comparing the one-sided limits, f'(π/2⁻) = -1 while f'(π/2⁺) = 1. Since the left-hand derivative does not equal the right-hand derivative (LHD ≠ RHD), the limit defining the derivative does not exist.
Conclusion
Because the left-hand derivative (-1) is unequal to the right-hand derivative (1), f(x) = |cos x| + |sin x| is not differentiable at x = π/2. Geometrically, the function exhibits a sharp corner at x = π/2 caused by the non-smooth sign reversal of cos x across zero.