Limits, Continuity & Differentiability
Differentiation
Grade 12
Question:
<p>If \(2y = \left(\cot^{-1}\left(\dfrac{\sqrt{3}\cos x + \sin x}{\cos x - \sqrt{3}\sin x}\right)\right)^2\), \(x \in \left(0, \dfrac{\pi}{2}\right)\), then \(\dfrac{dy}{dx}\) is equal to:</p>
<p>\(\dfrac{\pi}{6} - x\)</p>
<p>\(x - \dfrac{\pi}{6}\)</p>
<p>\(\dfrac{\pi}{3} - x\)</p>
<p>\(2x - \dfrac{\pi}{3}\)</p>
Step-by-Step Solution
Key Concept: Convert the cotangent inverse argument using the tangent addition formula: tan(A-B) = (tan A - tan B)/(1 + tan A·tan B), recognizing that cot⁻¹(u) = tan⁻¹(1/u). Then simplify using tan(x + π/3) to reduce the inverse trigonometric expression to a linear function.
<p><strong>Step 1: Simplify the argument of cot⁻¹</strong></p><p>Let u = (√3cos x + sin x)/(cos x - √3sin x). Divide numerator and denominator by cos x:</p><p>u = (√3 + tan x)/(1 - √3tan x)</p><p><strong>Step 2: Recognize the tangent addition formula</strong></p><p>Note that tan(π/3) = √3, so:</p><p>u = (tan(π/3) + tan x)/(1 - tan(π/3)·tan x) = tan(π/3 + x)</p><p><strong>Step 3: Apply cot⁻¹ property</strong></p><p>cot⁻¹(tan(π/3 + x)) = cot⁻¹(cot(π/2 - π/3 - x)) = cot⁻¹(cot(π/6 - x))</p><p>For x ∈ (0, π/2): cot⁻¹(cot(π/6 - x)) = π/6 - x (adjusting for domain)</p><p>Actually, using cot⁻¹(tan θ) = π/2 - θ:</p><p>cot⁻¹(tan(π/3 + x)) = π/2 - (π/3 + x) = π/6 - x</p><p><strong>Step 4: Differentiate</strong></p><p>2y = (π/6 - x)²</p><p>2(dy/dx) = 2(π/6 - x)·(-1) = -2(π/6 - x)</p><p>dy/dx = -(π/6 - x) = x - π/6</p><p>∴ Answer: C</p>
Correct Answer: C