Limits, Continuity & Differentiability
Differentiation using chain rule
Grade 12
Question:
<p>If \(t = e^{x^2}\) and \(y = t^2 - 1\) then \(\left(\dfrac{dy}{dx}\right)_{x=1}\) is</p>
<p>(a) \(\dfrac{1}{2e^2}\)</p>
<p>(b) \(2e\)</p>
<p>(c) 2</p>
<p>(d) \(2e^3\)</p>
Step-by-Step Solution
Key Concept: Use the chain rule by recognizing that y depends on x through an intermediate variable t. Compute dy/dt and dt/dx separately, then multiply them together.
<p><strong>Step 1:</strong> Identify the composite function structure.</p><p>Given: t = e^(x²) and y = t² - 1</p><p>y is a function of t, and t is a function of x, so use chain rule: dy/dx = (dy/dt) · (dt/dx)</p><p><strong>Step 2:</strong> Find dy/dt.</p><p>y = t² - 1</p><p>dy/dt = 2t</p><p><strong>Step 3:</strong> Find dt/dx.</p><p>t = e^(x²)</p><p>dt/dx = e^(x²) · 2x = 2x·e^(x²)</p><p><strong>Step 4:</strong> Apply chain rule.</p><p>dy/dx = (dy/dt) · (dt/dx) = 2t · 2x·e^(x²)</p><p>Since t = e^(x²):</p><p>dy/dx = 2·e^(x²) · 2x·e^(x²) = 4x·e^(2x²)</p><p><strong>Step 5:</strong> Evaluate at x = 1.</p><p>(dy/dx)ₓ₌₁ = 4(1)·e^(2·1²) = 4e²</p><p>∴ Answer: D</p>
Correct Answer: D