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Differentiability of |sin(x)| at x = 0

Consider the function f(x)=∣sin⁡(x)∣f(x) = |\sin(x)|. Its graph touches the x-axis at x=0x = 0 and x=πx = \pi (among other points), where the sine function changes sign.

Differentiability of |sin(x)| at x = 0 graph

Using the limit definition of the derivative, determine whether f(x)=∣sin⁡(x)∣f(x) = |\sin(x)| is differentiable at x=0x = 0. To justify your answer, compute the one-sided limits lim⁡h→0+f(0+h)−f(0)handlim⁡h→0−f(0+h)−f(0)h.\lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} \quad \text{and} \quad \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h}. State whether ff is differentiable at x=0x=0, and if not, describe what feature of the graph (from the list: corner, cusp, vertical tangent, discontinuity) causes the failure.

Hint

Near x=0x=0, sin⁡(h)\sin(h) is positive for small h>0h>0 and negative for small h<0h<0, so ∣sin⁡(h)∣|\sin(h)| splits into two different expressions on each side. Compute the difference quotient separately for h→0+h \to 0^+ and h→0−h \to 0^- using lim⁡h→0sin⁡hh=1\lim_{h\to 0}\frac{\sin h}{h}=1.

Step-by-step solution

For h>0h>0 small, sin⁡(h)>0\sin(h) > 0 so f(0+h)=∣sin⁡(h)∣=sin⁡(h)f(0+h) = |\sin(h)| = \sin(h), and since f(0)=0f(0)=0 the difference quotient is sin⁡(h)h\frac{\sin(h)}{h}, which tends to 11 as h→0+h \to 0^+. For h<0h<0 small, sin⁡(h)<0\sin(h) < 0 so f(0+h)=∣sin⁡(h)∣=−sin⁡(h)f(0+h) = |\sin(h)| = -\sin(h), giving the difference quotient −sin⁡(h)h\frac{-\sin(h)}{h}, which tends to −1-1 as h→0−h \to 0^-.

Since the right-hand limit (11) and left-hand limit (−1-1) of the difference quotient disagree, the overall limit does not exist, so ff is not differentiable at x=0x=0. Graphically, this matches the sharp corner in ∣sin⁡(x)∣|\sin(x)| at x=0x=0, analogous to the corner in ∣x∣|x| at 00.

Differentiability of |sin(x)| at x = π

Using the limit definition of the derivative, determine whether f(x)=∣sin⁡(x)∣f(x) = |\sin(x)| is differentiable at x=πx = \pi. Compute the one-sided limits lim⁡h→0+f(π+h)−f(π)handlim⁡h→0−f(π+h)−f(π)h\lim_{h \to 0^+} \frac{f(\pi+h) - f(\pi)}{h} \quad \text{and} \quad \lim_{h \to 0^-} \frac{f(\pi+h) - f(\pi)}{h} using the fact that sin⁡(π+h)=−sin⁡(h)\sin(\pi + h) = -\sin(h). State whether ff is differentiable at x=πx=\pi, and if not, identify the graphical feature responsible.

Hint

Use the identity sin⁡(π+h)=−sin⁡(h)\sin(\pi+h) = -\sin(h) to rewrite f(π+h)=∣sin⁡(π+h)∣=∣sin⁡(h)∣f(\pi+h) = |\sin(\pi+h)| = |\sin(h)|, then analyze the sign of sin⁡(h)\sin(h) for small positive and negative hh, just as in the x=0x=0 case.

Step-by-step solution

Since f(π)=∣sin⁡(π)∣=0f(\pi)=|\sin(\pi)|=0 and sin⁡(π+h)=−sin⁡(h)\sin(\pi+h) = -\sin(h), we have f(π+h)=∣−sin⁡(h)∣=∣sin⁡(h)∣f(\pi+h) = |-\sin(h)| = |\sin(h)|. So the difference quotient is ∣sin⁡(h)∣h\frac{|\sin(h)|}{h}, exactly the same expression analyzed at x=0x=0.

For h→0+h \to 0^+, ∣sin⁡(h)∣=sin⁡(h)|\sin(h)| = \sin(h), giving limit 11; for h→0−h \to 0^-, ∣sin⁡(h)∣=−sin⁡(h)|\sin(h)| = -\sin(h), giving limit −1-1. The one-sided limits disagree, so ff is not differentiable at x=πx = \pi either — again a corner point on the graph, just like at x=0x=0.

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