Limits, Continuity & Differentiability
Chain Rule and Composite Functions
Grade 12
Question:
<p><strong>Question 23:</strong> Given <p>We have <p><strong>f: (-1, 1) → ℝ</strong></p><p><strong>f(0) = -1, f'(0) = 1</strong></p><p>If <strong>g(x) = [f(2f(x) + 2)]^(1/2)</strong></p><p>Find <strong>g'(0)</strong></p>
Step-by-Step Solution
Key Concept: Apply the chain rule carefully to composite functions, identifying all intermediate derivatives needed.
<p><strong>Step 1:</strong> We have <strong>g(x) = [f(2f(x) + 2)]^(1/2)</strong>.</p><p><strong>Step 2:</strong> Differentiating using the chain rule: <strong>g'(x) = 2[f(2f(x) + 2)] · f'(2f(x) + 2) · 2f'(x)</strong>.</p><p><strong>Step 3:</strong> At x = 0: <strong>g'(0) = 2[f(2f(0) + 2)] · f'(2f(0) + 2) · 2f'(0)</strong>.</p><p><strong>Step 4:</strong> Substituting values: <strong>g'(0) = 2[f(0)] · f'(0) · 2f'(0) = 2(-1) · 1 · 2(1) = -4</strong>.</p><p>∴ The answer is <strong>-4</strong>.</p>
Correct Answer: -4