Limits, Continuity & Differentiability
Inverse Function Derivative
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: Use the derivative formula for inverse functions: $(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$, then differentiate again using chain rule to find the second derivative of the inverse function.
<p><strong>Step 1:</strong> Since $f(0) = 3$, we have $f^{-1}(3) = 0$.</p><p><strong>Step 2:</strong> For inverse functions: $(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$</p><p><strong>Step 3:</strong> Differentiate using chain rule to find the second derivative:</p><p>$(f^{-1})''(y) = -\frac{f''(f^{-1}(y)) \cdot (f^{-1})'(y)}{[f'(f^{-1}(y))]^2}$</p><p><strong>Step 4:</strong> Substitute $(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$:</p><p>$(f^{-1})''(y) = -\frac{f''(f^{-1}(y))}{[f'(f^{-1}(y))]^3}$</p><p><strong>Step 5:</strong> At $y = 3$: $(f^{-1})''(3) = -\frac{f''(0)}{[f'(0)]^3}$</p><p><strong>Step 6:</strong> Calculate $f'(0) = \sin^2(\sin(1))$ and $f''(0) = 2\sin(\sin(1))\cos(\sin(1))\cos(1)$</p><p><strong>Step 7:</strong> Simplify: $(f^{-1})''(3) = -\frac{2\sin(\sin(1))\cos(\sin(1))\cos(1)}{\sin^6(\sin(1))} = -\frac{2\cos(\sin(1))\cos(1)}{\sin^5(\sin(1))}$</p><p>∴ Answer: B</p>
Correct Answer: B