Limits, Continuity & Differentiability
Differentiation of Inverse Trigonometric Functions
Grade None
Question:
<p>If \(y = \sec(\tan^{-1} x)\), then \(\dfrac{dy}{dx}\) at \(x = 1\) is equal to:</p>
<p>\(\dfrac{1}{2}\)</p>
<p>\(\dfrac{1}{\sqrt{2}}\)</p>
<p>\(1\)</p>
<p>\(\sqrt{2}\)</p>
Step-by-Step Solution
Key Concept: Use the substitution θ = tan⁻¹(x) to convert the trigonometric composition into an algebraic form, then leverage the right triangle relationship where tan(θ) = x to find sec(θ) = √(1+x²).
<p><strong>Step 1:</strong> Let θ = tan⁻¹(x), so tan(θ) = x and we need to find sec(θ).</p><p><strong>Step 2:</strong> From the right triangle with tan(θ) = x/1, we have opposite = x and adjacent = 1, so hypotenuse = √(1+x²).</p><p><strong>Step 3:</strong> Therefore, sec(θ) = hypotenuse/adjacent = √(1+x²).</p><p><strong>Step 4:</strong> This gives us y = √(1+x²) = (1+x²)^(1/2).</p><p><strong>Step 5:</strong> Differentiate: dy/dx = (1/2)(1+x²)^(-1/2) · 2x = x/√(1+x²).</p><p><strong>Step 6:</strong> At x = 1: dy/dx|ₓ₌₁ = 1/√(1+1) = 1/√2 = √2/2.</p><p>∴ Answer: B</p>
Correct Answer: B