Differentiating Integrals

In Exercises 75–78, find dy/dx.
________
y = ∫₂ˣ √ 2 + cos³t dt

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Identify the function y as a definite integral with a variable upper limit: \(y = \int_{2}^{x} \sqrt{2 + \cos^{3} t} \, dt\).

Recall the Fundamental Theorem of Calculus Part 1, which states that if \(y = \int_{a}^{x} f(t) \, dt\), then \(\frac{dy}{dx} = f(x)\), provided \(f\) is continuous.

Apply the theorem directly to the given integral, recognizing that the integrand is \(f(t) = \sqrt{2 + \cos^{3} t}\).

Write the derivative as \(\frac{dy}{dx} = \sqrt{2 + \cos^{3} x}\), substituting the upper limit \(x\) into the integrand.

No further simplification is necessary unless specified; this expression represents the derivative \(\frac{dy}{dx}\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Fundamental Theorem of Calculus Part 1

This theorem connects differentiation and integration by stating that if y is defined as an integral with a variable upper limit, then dy/dx equals the integrand evaluated at that upper limit. It allows us to differentiate integrals with variable limits directly.

Chain Rule

When the upper limit of the integral is a function of x (not just x itself), the chain rule is used to differentiate the integral. It involves multiplying the derivative of the integral's upper limit by the integrand evaluated at that limit.

Integrand Function and Its Evaluation

Understanding the integrand, here √(2 + cos³t), is crucial because it must be evaluated at the upper limit of integration after differentiation. Recognizing how to substitute the variable limit into the integrand is key to finding dy/dx.

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