Applications of Derivatives
Inverse Functions and Derivatives
Grade 12
Question:
<p><strong>Paragraph for Questions 632 and 633</strong><br>Suppose that \(f\) is defined on \(R\) by the rule \(f(x) = (1-x)(1+x^2)\). The function is invertible and its inverse is denoted by \(f^{-1}\).</p><p>If \(h = f^{-1}(\ln(f(x)))\), \(x < 1\), then the value of \(\left(3 + \dfrac{1}{h'(0)}\right)\) is:</p>
<p>(a) 2</p>
<p>(b) 3</p>
<p>(c) 5</p>
<p>(d) 6</p>
Step-by-Step Solution
Key Concept: To find dh/dx, use the chain rule: dh/dx = d/dx[f⁻¹(ln(f(x)))] requires differentiating f⁻¹ using the formula (f⁻¹)'(y) = 1/f'(f⁻¹(y)), then multiply by the derivative of ln(f(x)) which is f'(x)/f(x).
<p><strong>Step 1:</strong> Given h = f⁻¹(ln(f(x))), differentiate using chain rule:</p><p>dh/dx = [1/f'(f⁻¹(ln(f(x))))] · d/dx[ln(f(x))]</p><p><strong>Step 2:</strong> Compute d/dx[ln(f(x))] = f'(x)/f(x)</p><p><strong>Step 3:</strong> Therefore: dh/dx = [f'(x)/f(x)] · [1/f'(f⁻¹(ln(f(x))))]</p><p><strong>Step 4:</strong> Since h = f⁻¹(ln(f(x))), we have f(h) = ln(f(x)), so f⁻¹(ln(f(x))) = h</p><p><strong>Step 5:</strong> Substitute: dh/dx = [f'(x)/f(x)] · [1/f'(h)]</p><p>∴ Answer: C</p>
Correct Answer: C