Find dy/dx if y = ∫ₓ¹ √(1 + t²)dt.

Explain the main steps in your calculation.

Guida verificata passo dopo passo

Recognize that the function y is defined as a definite integral with a variable lower limit of integration: \(y = \int_{x}^{1} \sqrt{1 + t^{2}} \, dt\).

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

In this problem, the upper limit is a constant (1), so \(b(x) = 1\) and \(b'(x) = 0\). The lower limit is \(a(x) = x\), so \(a'(x) = 1\).

Apply the Leibniz rule: \(\frac{dy}{dx} = 0 - \sqrt{1 + x^{2}} \cdot 1 = -\sqrt{1 + x^{2}}\).

Thus, the derivative \(\frac{dy}{dx}\) is the negative of the integrand evaluated at the lower limit \(x\).

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Fundamental Theorem of Calculus

This theorem connects differentiation and integration, stating that if a function is defined as an integral with a variable limit, its derivative can be found by evaluating the integrand at that limit. Specifically, if y = ∫ₐˣ f(t) dt, then dy/dx = f(x).

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Fundamental Theorem of Calculus Part 1

Leibniz Rule for Differentiation under the Integral Sign

When the limits of integration are functions of x, the derivative of the integral involves evaluating the integrand at the limits multiplied by the derivatives of those limits. For y = ∫_{g(x)}^{h(x)} f(t) dt, dy/dx = f(h(x))·h'(x) - f(g(x))·g'(x).

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Additional Rules for Indefinite Integrals

Chain Rule

The chain rule is used to differentiate composite functions. In this problem, since the upper or lower limit of integration is a function of x, applying the chain rule helps find the derivative of the integral by differentiating the limits and the integrand accordingly.

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