Limits, Continuity & Differentiability
Differentiation of inverse trigonometric functions
Grade 12
Question:
<p>If \(f(t) = \tan^{-1}\left[\dfrac{1}{2}(\sqrt{1+t^2}-1)\right]\) then \(f'(0)\) is</p>
<p>(a) \(\dfrac{1}{2}\)</p>
<p>(b) 0</p>
<p>(c) 1</p>
<p>(d) does not exist</p>
Step-by-Step Solution
Key Concept: Use the chain rule combined with the derivative of tan⁻¹(u) which is 1/(1+u²)·u'. The inner expression √(1+t²)-1 can be simplified using rationalization or direct differentiation.
<p><strong>Step 1:</strong> Let u(t) = ½(√(1+t²)-1). Then f(t) = tan⁻¹(u(t)).</p><p><strong>Step 2:</strong> By chain rule: f'(t) = 1/(1+u²) · u'(t)</p><p><strong>Step 3:</strong> Find u'(t): u'(t) = ½ · d/dt[√(1+t²)] = ½ · 2t/(2√(1+t²)) = t/(2√(1+t²))</p><p><strong>Step 4:</strong> At t=0: u(0) = ½(√1 - 1) = 0, so 1+u²(0) = 1</p><p><strong>Step 5:</strong> u'(0) = 0/(2√1) = 0</p><p><strong>Step 6:</strong> Therefore f'(0) = 1/(1+0) · 0 = <strong>0</strong></p><p>∴ Answer: A</p>
Correct Answer: A