In Exercises 75–78, find dy/dx.
y = ∫(from x to 1) (6/(3 + t^4))dt

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Recognize that the function y is defined as a definite integral with a variable limit: \(y = \int_{x}^{1} \frac{6}{3 + t^{4}} \, dt\).

Recall the Leibniz rule for differentiation of an integral with variable limits: if \(y = \int_{a(x)}^{b(x)} f(t) \, dt\), then \(\frac{dy}{dx} = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)\).

Identify the limits of integration: the lower limit is \(x\) and the upper limit is the constant \(1\). Therefore, \(a(x) = x\) and \(b(x) = 1\).

Calculate the derivatives of the limits: \(a'(x) = \frac{d}{dx} x = 1\) and \(b'(x) = \frac{d}{dx} 1 = 0\).

Apply the Leibniz rule: \(\frac{dy}{dx} = f(1) \cdot 0 - f(x) \cdot 1 = - \frac{6}{3 + x^{4}}\).

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주요 개념

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Fundamental Theorem of Calculus Part 1

This theorem connects differentiation and integration, stating that if a function is defined as an integral with a variable limit, its derivative is the integrand evaluated at that limit, multiplied by the derivative of the limit. It allows us to differentiate integrals with variable limits directly.

Leibniz Rule for Differentiation under the Integral Sign

Leibniz Rule generalizes differentiation of integrals with variable limits. When both limits depend on x, the derivative of the integral is the integrand evaluated at the upper limit times the derivative of the upper limit minus the integrand at the lower limit times the derivative of the lower limit.

Chain Rule

The chain rule is used to differentiate composite functions. When the limits of integration are functions of x, the derivative of the integral involves multiplying the integrand evaluated at the limit by the derivative of that limit, applying the chain rule to handle the inner function.

In Exercises 75–78, find dy/dx.y = ∫(from x to 1) (6/(3 + t^4))dt