Find dy/dx if y = ∫ₓ¹ √(1 + t²)dt.

Explain the main steps in your calculation.

Verified step by step guidance

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 \(x\).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

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

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).

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).

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.

Recommended video: