Limits, Continuity & Differentiability
Differentiation of Parametric Functions
Grade 12
Question:
<p>If \(x = \sqrt{2^{\sec^{-1}t}}\) and \(y = \sqrt{2^{\csc^{-1}t}}\) \((|t| \geq 1)\), then \(\dfrac{dy}{dx}\) is equal to</p>
<p>\(\dfrac{y}{x}\)</p>
<p>\(\dfrac{x}{y}\)</p>
<p>\(-\dfrac{y}{x}\)</p>
<p>\(-\dfrac{x}{y}\)</p>
Step-by-Step Solution
Key Concept: Express y and x in terms of inverse trigonometric functions, then use the relationship sec⁻¹t + csc⁻¹t = π/2 to find dy/dx through logarithmic differentiation or direct substitution.
<p><strong>Step 1:</strong> Use the identity sec⁻¹t + csc⁻¹t = π/2, so csc⁻¹t = π/2 - sec⁻¹t</p><p><strong>Step 2:</strong> Express y in terms of sec⁻¹t:</p><p>y = √(2^(csc⁻¹t)) = √(2^(π/2 - sec⁻¹t)) = √(2^(π/2) · 2^(-sec⁻¹t)) = √(2^(π/2)) · √(2^(-sec⁻¹t))</p><p><strong>Step 3:</strong> Let u = sec⁻¹t. Then x = √(2^u) = 2^(u/2) and y = √(2^(π/2)) · 2^(-u/2) = 2^(π/4) · 2^(-u/2)</p><p><strong>Step 4:</strong> Taking logarithms: ln(x) = (u/2)ln(2) and ln(y) = (π/4)ln(2) - (u/2)ln(2)</p><p><strong>Step 5:</strong> Differentiate with respect to t:</p><p>1/x · dx/dt = (1/2)ln(2) · du/dt and 1/y · dy/dt = -(1/2)ln(2) · du/dt</p><p><strong>Step 6:</strong> Therefore: dy/dx = (1/y · dy/dt)/(1/x · dx/dt) = -1</p><p>∴ Answer: C</p>
Correct Answer: C