Question

If \(y=\int_{0}^{\sqrt{x}} \sin t^{2} d t,\) find \(d y / d x.\)

Short answer

\(\frac{dy}{dx} = \sin(x) \cdot \frac{1}{2\sqrt{x}}\)

Step-by-step solution

Identify the integral expression

The given function is: \[ y = \int_{0}^{\sqrt{x}} \sin{t^2} \, dt \]

Recognize the Fundamental Theorem of Calculus Part 2 and use Leibniz's Rule

We need to use the Fundamental Theorem of Calculus Part 2 and Leibniz's Rule to differentiate the integral with respect to \(x\). According to these rules, if \(y = \int_{a(x)}^{b(x)} f(t) \, dt\), then \ \[ \frac{d}{dx} y = f(b(x)) \cdot \frac{d}{dx} b(x) - f(a(x)) \cdot \frac{d}{dx} a(x). \]

Apply the rules to the given problem

In our problem, \(a(x) = 0\) and \(b(x) = \sqrt{x}\). Therefore, \[ \frac{d}{dx} y = \sin(\sqrt{x}^2) \cdot \frac{d}{dx} (\sqrt{x}) - \sin(0^2) \cdot \frac{d}{dx} (0). \]

Evaluate the derivatives and simplify

Since \(\frac{d}{dx} (0) = 0\), we get: \[ \frac{d}{dx} y = \sin(x) \cdot \frac{d}{dx} (\sqrt{x}) - 0. \]Also, \(\frac{d}{dx} (\sqrt{x}) = \frac{1}{2\sqrt{x}}\). Substituting this back, we have: \[ \frac{d}{dx} y = \sin(x) \cdot \frac{1}{2\sqrt{x}}. \]

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, showing they are essentially inverse processes. It has two parts. The second part of the theorem is crucial for our work here. It states that if we have an integral function like \( F(x) = \int_a^x f(t) \, dt \), the derivative of this integral with respect to \(x\) is simply the integrand evaluated at \(x\), i.e., \( F'(x) = f(x) \). This means that differentiation undoes the integration. This theorem simplifies our work when finding the derivative of integral-based functions.

Leibniz's Rule

Leibniz's Rule is a powerful tool for differentiating integral expressions where the limits depend on the variable. Say we have \( y = \int_{a(x)}^{b(x)} f(t) \, dt \), the rule helps us understand how to find the derivative. According to Leibniz's Rule, the derivative is:
\[ \frac{d}{dx} y = f(b(x)) \cdot \frac{d}{dx} b(x) - f(a(x)) \cdot \frac{d}{dx} a(x). \]

• First term: Evaluate the integrand function \( f(t)\) at the upper limit \( b(x) \) and multiply by the derivative of this limit with respect to \(x\).
• Second term: Evaluate the integrand function \( f(t) \) at the lower limit \( a(x) \) and multiply by the derivative of this limit with respect to \(x\). However, in our problem, \( a(x) = 0 \) so its derivative is zero.

This approach helps us break down complex integrals step by step.

Derivative of Integral

Using the previous concepts, we can now find the derivative of the given function \( y = \int_{0}^{\sqrt{x}} \sin(t^2) \, dt \). Here's how we apply Leibniz's Rule:

• Upper limit derivative: \( b(x) = \sqrt{x} \), its derivative is \( \frac{d}{dx} \sqrt{x} = \frac{1}{2\sqrt{x}} \).
• Evaluate the integrand: Substitute the upper limit \( \sqrt{x} \) into \( \sin(t^2) \). So we get \( \sin((\sqrt{x})^2) = \sin(x) \).
• Result: Multiply \( \sin(x) \) by \( \frac{1}{2\sqrt{x}} \), giving \( \[ \frac{d}{dx} y = \sin(x) \cdot \frac{1}{2\sqrt{x}} \). \]

Therefore, the derivative of \( y \) with respect to \( x \) is \( \frac{d y}{d x} = \sin(x) \cdot \frac{1}{2\sqrt{x}} \). Noticeably, the process highlights how the Fundamental Theorem and Leibniz's Rule simplify the differentiation of integrals effectively.