Limits, Continuity & Differentiability
Differentiability of composite functions
Grade 12
Question:
<p>Let <i>f</i>(<i>x</i>) = <i>x</i>|<i>x</i>|, <i>g</i>(<i>x</i>) = sin <i>x</i> and <i>h</i>(<i>x</i>) = (<i>g</i> ∘ <i>f</i>)(<i>x</i>). Then</p>
<p><i>h</i>(<i>x</i>) is not differentiable at <i>x</i> = 0.</p>
<p><i>h</i>(<i>x</i>) is differentiable at <i>x</i> = 0, but <i>h'</i>(<i>x</i>) is not continuous at <i>x</i> = 0.</p>
<p><i>h'</i>(<i>x</i>) is continuous at <i>x</i> = 0 but it is not differentiable at <i>x</i> = 0.</p>
<p><i>h'</i>(<i>x</i>) is differentiable at <i>x</i> = 0.</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) = x|x| is an odd function with continuous derivative everywhere, and composition with sin preserves differentiability. The key is checking h'(0) exists by verifying the derivative from first principles at the critical point.
<p><strong>Step 1:</strong> Find h(x) = (g ∘ f)(x) = sin(f(x)) = sin(x|x|)</p><p><strong>Step 2:</strong> Analyze f(x) = x|x|: This equals x² when x≥0 and -x² when x<0. So f'(x) = 2x for x≥0 and f'(x) = -2x for x<0, giving f'(0) = 0. Thus f is differentiable everywhere.</p><p><strong>Step 3:</strong> Since f is differentiable everywhere and g(x) = sin x is differentiable everywhere, their composition h(x) = sin(x|x|) is differentiable everywhere by the chain rule.</p><p><strong>Step 4:</strong> Calculate h'(x) = cos(x|x|) · f'(x) = cos(x|x|) · 2|x|. At x=0: h'(0) = cos(0) · 0 = 0.</p><p><strong>Step 5:</strong> h is continuous everywhere (composition of continuous functions) and differentiable everywhere (composition of differentiable functions).</p><p>∴ Answer: D (h is both continuous and differentiable everywhere)</p>
Correct Answer: D