Limits, Continuity & Differentiability
Differentiability of composite functions
Grade 12
Question:
<p>Let \(f(x) = x|x|\) and \(g(x) = \sin x\). Then \(g \circ f\) is:</p>
<p>Differentiable everywhere except at \(x = 0\)</p>
<p>Not differentiable at \(x = 0\)</p>
<p>Differentiable everywhere</p>
<p>Not continuous at \(x = 0\)</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) = x|x| is differentiable everywhere (including at x=0) because its derivative exists from both sides, and composition of differentiable functions preserves differentiability. The absolute value in f doesn't destroy differentiability at the origin.
<p><strong>Step 1:</strong> Analyze f(x) = x|x|. Rewrite as: f(x) = x² when x ≥ 0, and f(x) = -x² when x < 0.</p><p><strong>Step 2:</strong> Check differentiability of f at x=0. Left derivative: lim(h→0⁻) [f(h)-f(0)]/h = lim(h→0⁻) [-h²]/h = 0. Right derivative: lim(h→0⁺) [h²]/h = 0. Since both equal 0, f'(0) = 0 exists.</p><p><strong>Step 3:</strong> For x ≠ 0: f'(x) = 2|x|, which is continuous everywhere including at x=0 (where it equals 0).</p><p><strong>Step 4:</strong> Now g(f(x)) = sin(x|x|). Since f is differentiable everywhere and g = sin is differentiable everywhere, their composition is differentiable everywhere by the chain rule.</p><p><strong>Step 5:</strong> Therefore (g∘f) is differentiable on ℝ (which implies continuous and has all derivatives).</p><p>∴ Answer: C</p>
Correct Answer: C