Problem 31 - Calculus for Biology and Medicine

Question

Use Leibniz's rule to find \(\frac{d y}{d x}\). $$ y=\int_{x^{2}}^{1} \sec t d t,-1

Step-by-Step Solution

Answer

\(\frac{d y}{d x} = -2x \sec(x^2)\).

Step 1: Understand Leibniz's Rule

Leibniz's rule is used to differentiate an integral whose limits are functions of the variable you want to differentiate with respect to. It is given by the formula:\[\frac{d}{dx} \left( \int_{a(x)}^{b(x)} f(t, x) \, dt \right) = f(b(x), x) \cdot b'(x) - f(a(x), x) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(t, x) \, dt\]In this problem, there is no explicit dependence of the integrand \(\sec t\) on \(x\), so the partial derivative term is zero.

Step 2: Identify the Limits of Integration

The limits of integration are given as \(a(x) = x^{2}\) and \(b(x) = 1\). These will be used to apply Leibniz's rule.

Step 3: Calculate Derivatives of Limits

Compute the derivatives of the upper and lower limits with respect to \(x\):- The derivative of the upper limit \(b(x) = 1\) is \(b'(x) = 0\).- The derivative of the lower limit \(a(x) = x^{2}\) is \(a'(x) = 2x\).

Step 4: Apply Leibniz's Rule

Using Leibniz's rule, the derivative \(\frac{d y}{d x}\) is given by:\[\frac{d y}{d x} = \sec(1(x)) \cdot 0 - \sec(x^2) \cdot 2x + \int_{x^2}^{1} 0 \, dt\]This simplifies to:\[\frac{d y}{d x} = -2x \sec(x^2)\]

Step 5: Simplification and Final Result

Since the integral term is zero (as noted in Step 1), the derivative simplifies to \[\frac{d y}{d x} = -2x \sec(x^2)\]This is our final result for the derivative.

Key Concepts

Leibniz's Rule

Leibniz's Rule is a powerful tool in calculus for handling the differentiation of integrals with variable limits. When faced with an integral whose limits are functions involving the variable of differentiation, this rule helps find the derivative efficiently. Here's the rule in simple terms: If you have an integral of the form \[ \int_{a(x)}^{b(x)} f(t, x) \, dt \] and you want to differentiate it with respect to \( x \), Leibniz’s rule states:\[ \frac{d}{dx} \left( \int_{a(x)}^{b(x)} f(t, x) \, dt \right) = f(b(x), x) \cdot b'(x) - f(a(x), x) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(t, x) \, dt \]In the formula:

• \( f(t, x) \) is the integrand.
• \( a(x) \) and \( b(x) \) are the lower and upper limits.
• \( a'(x) \) and \( b'(x) \) are the derivatives of the limits with respect to \( x \).
• \( \frac{\partial}{\partial x} f(t, x) \) is the partial derivative of \( f \) with respect to \( x \).

In our specific exercise, since \( \sec t \) does not explicitly depend on \( x \), the last integral term is zero.

Integration Limits

Integration limits are the start and end points of an integral. In Leibniz’s Rule, these limits can be functions involving the variable of differentiation, making it crucial to understand their behavior when applying the rule.For the exercise, our limits are \( a(x) = x^{2} \) and \( b(x) = 1 \). The limits serve two purposes:

• They define the interval over which the function is integrated.
• They often change with respect to \( x \), and their derivatives with respect to \( x \) need to be calculated.

By differentiating these limits:

• The derivative of the upper limit, \( b(x) = 1 \), is \( b'(x) = 0 \).
• The derivative of the lower limit, \( a(x) = x^{2} \), is \( a'(x) = 2x \).

These derivatives are then plugged into Leibniz’s Rule to help calculate the derivative of the integral with respect to \( x \).

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative represents the rate of change of the function with respect to its variables. In calculus, especially with Leibniz's rule applications, differentiation helps determine how integrals change as their limits vary.To differentiate the integral from our exercise:We apply Leibniz’s Rule:1. Calculate the function value at the upper limit, \( \, b(x) = 1 \), but since \( b'(x) = 0 \), this term becomes zero.2. Calculate the function value at the lower limit, \( f(x^2, x) = \sec(x^2) \), which results in the term \(-2x \sec(x^2) \) (utilizing \( a'(x) = 2x \)).3. Recognize that the integral term vanishes because the partial derivative \( \frac{\partial}{\partial x} (\sec t) \) is zero.Thus, the derivative for our problem is simplified to:\[ \frac{d y}{d x} = -2x \sec(x^2) \]This reflects how the integral of \( \sec t \) between the limits \( x^2 \) and \( 1 \) changes with \( x \). Differentiating not only brings out the rate of change but also highlights the behavior of the function within the specified limits.