Limits, Continuity & Differentiability
Implicit Differentiation
Grade 12
Question:
<p>If \(e^y + xy = e\), the ordered pair \(\left(\dfrac{dy}{dx}, \dfrac{d^2y}{dx^2}\right)\) at \(x = 0\) is equal to:</p>
<p>\(\left(\dfrac{1}{e}, -\dfrac{1}{e^2}\right)\)</p>
<p>\(\left(-\dfrac{1}{e}, \dfrac{1}{e^2}\right)\)</p>
<p>\(\left(\dfrac{1}{e}, \dfrac{1}{e^2}\right)\)</p>
<p>\(\left(-\dfrac{1}{e}, -\dfrac{1}{e^2}\right)\)</p>
Step-by-Step Solution
Key Concept: Use implicit differentiation on the constraint equation to find dy/dx, then differentiate again to find d²y/dx². First solve for y at x=0 using the original equation.
<p><strong>Step 1:</strong> Find y at x = 0. Substitute x = 0 into e^y + xy = e:<br>e^y + 0 = e ⟹ e^y = e ⟹ y = 1</p><p><strong>Step 2:</strong> Differentiate e^y + xy = e implicitly with respect to x:<br>e^y(dy/dx) + y + x(dy/dx) = 0<br>(e^y + x)(dy/dx) = -y<br>dy/dx = -y/(e^y + x)</p><p><strong>Step 3:</strong> Evaluate dy/dx at (0,1):<br>dy/dx|(0,1) = -1/(e + 0) = -1/e</p><p><strong>Step 4:</strong> Differentiate dy/dx = -y/(e^y + x) using quotient rule:<br>d²y/dx² = -[(e^y + x)(dy/dx) - y(e^y·dy/dx + 1)]/(e^y + x)²<br>d²y/dx² = -[(e^y + x)(dy/dx) - y(e^y·dy/dx + 1)]/(e^y + x)²</p><p><strong>Step 5:</strong> Substitute x = 0, y = 1, dy/dx = -1/e, e^y = e:<br>d²y/dx² = -[(e + 0)(-1/e) - 1(e·(-1/e) + 1)]/(e + 0)²<br>d²y/dx² = -[-1 - 1(-1 + 1)]/e²<br>d²y/dx² = -[-1 - 0]/e² = 1/e²</p><p>∴ Answer: D: (-1/e, 1/e²)</p>
Correct Answer: D