Limits, Continuity & Differentiability
Differentiation of inverse trigonometric functions
Grade 12
Question:
<p>Given: \(2y = \left(\cot^{-1}\left(\dfrac{\sqrt{3}\cos x + \sin x}{\cos x - \sqrt{3}\sin x}\right)\right)^2\). Find \(\dfrac{dy}{dx}\) (or simplify \(2y\)).</p>
<p>\(2y = x^2 - \dfrac{\pi}{3}x + \dfrac{\pi^2}{36}\)</p>
<p>\(2y = x^2 + \dfrac{\pi}{3}x + \dfrac{\pi^2}{36}\)</p>
<p>\(2y = x^2 - \dfrac{\pi}{6}x + \dfrac{\pi^2}{36}\)</p>
<p>\(2y = x^2 + \dfrac{\pi}{6}x + \dfrac{\pi^2}{36}\)</p>
Step-by-Step Solution
Key Concept: Convert the cotangent inverse argument using trigonometric identity: express √3cos x + sin x and cos x - √3sin x in terms of single angle functions, recognizing that cot⁻¹(√3cos x + sin x)/(cos x - √3sin x) reduces to a simple linear function of x after applying the tangent addition formula.
<p><strong>Step 1: Simplify the argument of cot⁻¹</strong></p><p>Rewrite numerator: √3cos x + sin x = 2[√3/2 cos x + 1/2 sin x] = 2sin(x + π/3)</p><p>Rewrite denominator: cos x - √3sin x = 2[1/2 cos x - √3/2 sin x] = 2sin(π/6 - x) = 2cos(x + π/3)</p><p><strong>Step 2: Simplify the cotangent inverse</strong></p><p>cot⁻¹(2sin(x + π/3)/2cos(x + π/3)) = cot⁻¹(tan(x + π/3)) = cot⁻¹(cot(π/2 - (x + π/3))) = π/2 - x - π/3 = π/6 - x</p><p><strong>Step 3: Express 2y</strong></p><p>2y = (π/6 - x)²</p><p><strong>Step 4: Differentiate</strong></p><p>dy/dx = 1/2 · 2(π/6 - x)·(-1) = -2(π/6 - x) = 2x - π/3</p><p>∴ <strong>2y = (π/6 - x)² or dy/dx = 2x - π/3</strong></p>
Correct Answer: A