Limits, Continuity & Differentiability
Parametric differentiation
Grade 12
<p>Given: \(x = \sqrt{2^{\cosec^{-1}t}}\) and \(y = \sqrt{2^{\sec^{-1}t}}\) where \(|t| \geq 1\). Find \(\dfrac{dy}{dx}\).</p>
Step-by-Step Solution
Key Concept: Use logarithmic differentiation and the chain rule on both parametric expressions, then recognize that dx/dt and dy/dt both depend on the derivative of inverse trigonometric functions (1/(t√(t²-1))), allowing simplification of dy/dx.
<p><strong>Step 1:</strong> Express x and y using logarithms for easier differentiation.</p><p>From x = √(2^(cosec⁻¹t)), we get: 2ln(x) = cosec⁻¹(t)</p><p>From y = √(2^(sec⁻¹t)), we get: 2ln(y) = sec⁻¹(t)</p><p><strong>Step 2:</strong> Differentiate both equations with respect to t.</p><p>For x: 2·(1/x)·(dx/dt) = -1/(|t|√(t²-1))</p><p>Therefore: dx/dt = -x/(2|t|√(t²-1))</p><p>For y: 2·(1/y)·(dy/dt) = 1/(|t|√(t²-1))</p><p>Therefore: dy/dt = y/(2|t|√(t²-1))</p><p><strong>Step 3:</strong> Calculate dy/dx = (dy/dt)/(dx/dt).</p><p>dy/dx = [y/(2|t|√(t²-1))] / [-x/(2|t|√(t²-1))]</p><p>dy/dx = y/(-x) = <strong>-y/x</strong></p><p><strong>Step 4:</strong> Express the answer in standard form.</p><p>Since y = √(2^(sec⁻¹t)) and x = √(2^(cosec⁻¹t)), the answer is <strong>-√(2^(sec⁻¹t - cosec⁻¹t))</strong> or equivalently <strong>-y/x</strong></p><p>∴ Answer: B</p>
Correct Answer: B