Limits, Continuity & Differentiability
Derivative of Inverse Function
Grade 12
Question:
<p>If \(f(x)\) is a real valued bijective function satisfying \(f'(x) = \sin^2(\sin(x+1))\) and \(f(0) = 3\), then the value of \((f^{-1})''(3)\) is equal to:</p>
<p>(a) \(-\dfrac{2\sin(\cos 1)\sin 1}{\sin^5(\cos 1)}\)</p>
<p>(b) \(-\dfrac{2\sin(\sin 1)\cos 1}{\sin^5(\sin 1)}\)</p>
<p>(c) \(-\dfrac{2\sin(\cos 1)\sin^2 1}{\sin^6(\cos 1)}\)</p>
<p>(d) \(-\dfrac{\sin^2(\sin 1)}{\cos^2(\cos 1)}\)</p>
Step-by-Step Solution
Key Concept: For inverse functions, use the relationship (f⁻¹)''(y) = -f''(x)/[f'(x)]³ where y = f(x). Since f(0) = 3, we have x = 0 when y = 3, so we need f'(0) and f''(0).
<p><strong>Step 1:</strong> Find f'(0).</p><p>f'(x) = sin²(sin(x+1))</p><p>f'(0) = sin²(sin(1))</p><p><strong>Step 2:</strong> Find f''(x) by differentiating f'(x).</p><p>f''(x) = 2sin(sin(x+1)) · cos(sin(x+1)) · cos(x+1)</p><p>f''(0) = 2sin(sin(1)) · cos(sin(1)) · cos(1)</p><p><strong>Step 3:</strong> Apply the inverse function second derivative formula.</p><p>For y = f(x), if f(0) = 3, then (f⁻¹)''(3) = -f''(0)/[f'(0)]³</p><p>(f⁻¹)''(3) = -2sin(sin(1))cos(sin(1))cos(1) / [sin²(sin(1))]³</p><p>(f⁻¹)''(3) = -2cos(sin(1))cos(1) / sin⁴(sin(1))</p><p>∴ Answer: B</p>
Correct Answer: B