Limits, Continuity & Differentiability
Differentiation using Chain Rule
Grade None
Question:
<p>Given \(f(1) = 1\), \(f'(1) = 3\). Find the derivative of \(f[f(f(x))] + (f(x))^2\) at \(x = 1\).</p>
Step-by-Step Solution
Key Concept: Apply the chain rule systematically for composite functions, then use the given values f(1)=1 and f'(1)=3 to evaluate at x=1. The key is recognizing that f(f(1))=f(1)=1 due to the fixed point structure.
<p><strong>Step 1:</strong> Let g(x) = f[f(f(x))] + (f(x))²</p><p><strong>Step 2:</strong> Find g'(x) using chain rule:</p><p>For f[f(f(x))]: d/dx[f[f(f(x))]] = f'[f(f(x))] · f'[f(x)] · f'(x)</p><p>For (f(x))²: d/dx[(f(x))²] = 2f(x) · f'(x)</p><p><strong>Step 3:</strong> Therefore, g'(x) = f'[f(f(x))] · f'[f(x)] · f'(x) + 2f(x) · f'(x)</p><p><strong>Step 4:</strong> Evaluate at x = 1:</p><p>Since f(1) = 1, we have f(f(1)) = f(1) = 1, and f(f(f(1))) = f(1) = 1</p><p>g'(1) = f'[f(f(1))] · f'[f(1)] · f'(1) + 2f(1) · f'(1)</p><p>g'(1) = f'(1) · f'(1) · f'(1) + 2(1) · f'(1)</p><p>g'(1) = [f'(1)]³ + 2f'(1)</p><p>g'(1) = (3)³ + 2(3) = 27 + 6 = 33</p><p>∴ <strong>Answer: 33</strong></p>
Correct Answer: 33