<p>If \(y = \int_0^{f(x)} \frac{dt}{1+t^4}\), find \(\frac{dy}{dx}\).</p>
Step-by-Step Solution
Key Concept: Use Leibniz integral rule for differentiation under the integral sign: when the upper limit is a function of x, multiply the integrand (evaluated at upper limit) by the derivative of that limit.
<p><strong>Step 1:</strong> Recognize that y is defined as a definite integral with variable upper limit f(x) and constant lower limit 0.</p><p><strong>Step 2:</strong> Apply Leibniz rule for differentiation under the integral sign: if y = ∫₀^(f(x)) g(t)dt, then dy/dx = g(f(x)) · f'(x)</p><p><strong>Step 3:</strong> Here g(t) = 1/(1+t⁴), so evaluate at upper limit: g(f(x)) = 1/(1+[f(x)]⁴)</p><p><strong>Step 4:</strong> Multiply by the derivative of the upper limit: dy/dx = 1/(1+[f(x)]⁴) · f'(x)</p><p>∴ Answer: f'(x)/(1+[f(x)]⁴)</p>
Correct Answer: f'(x)/(1+[f(x)]^4)