Problem 14 - Calculus: Early Transcendentals

Question

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. \( \displaystyle h(x) = \int^{\sqrt{x}}_1 \frac{z^2}{z^4 + 1} \,dz \)

Step-by-Step Solution

Answer

The derivative of the function is \( h'(x) = \frac{\sqrt{x}}{2(x^2 + 1)} \).

Step 1: Identify Part 1 of the Fundamental Theorem of Calculus

Part 1 of the Fundamental Theorem of Calculus states that if a function is defined as \[ F(x) = \int_{a(x)}^{b(x)} f(t) \, dt \]where \( f \) is continuous, then the derivative \( F'(x) \) is given by \[ F'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \] provided \( a \) and \( b \) are differentiable functions of \( x \).

Step 2: Identify the upper and lower limits and the integrand

In the given function:\[ h(x) = \int_{1}^{\sqrt{x}} \frac{z^2}{z^4 + 1} \, dz \]we have the following components:- Lower limit: \( a(x) = 1 \)- Upper limit: \( b(x) = \sqrt{x} \)- Integrand: \( f(z) = \frac{z^2}{z^4 + 1} \)

Step 3: Calculate the derivative of the upper and lower limits

Differentiate the upper and lower limits with respect to \( x \):- The derivative of the lower limit: \[ a'(x) = 0 \] because \( a(x) = 1 \) is constant.- The derivative of the upper limit: \[ b'(x) = \frac{d}{dx} (\sqrt{x}) = \frac{1}{2\sqrt{x}} \]

Step 4: Apply the Fundamental Theorem of Calculus

Using the Fundamental Theorem of Calculus Part 1, compute:\[ h'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \]Substitute the identified terms:\[ h'(x) = \left( \frac{(\sqrt{x})^2}{(\sqrt{x})^4 + 1} \right) \cdot \frac{1}{2\sqrt{x}} - \left( \frac{1^2}{1^4 + 1} \right) \cdot 0 \]This simplifies to:\[ h'(x) = \left( \frac{x}{x^2 + 1} \right) \cdot \frac{1}{2\sqrt{x}} \]

Step 5: Simplify the result

Simplify the expression for \( h'(x) \):\[ h'(x) = \frac{x}{x^2 + 1} \cdot \frac{1}{2\sqrt{x}} = \frac{x}{2\sqrt{x}(x^2 + 1)} \] Further simplify the fraction by combining terms:\[ h'(x) = \frac{\sqrt{x}}{2(x^2 + 1)} \]

Key Concepts

Derivative of an Integral

The Fundamental Theorem of Calculus beautifully connects differentiation and integration. Specifically, it tells us how to find the derivative of an integral like the one in our problem. This is all about using Part 1 of the theorem. Imagine you have a function defined as an integral with variable limits such as \(F(x) = \int_{a(x)}^{b(x)} f(t) \ dt\). This theorem then gives you a neat formula for its derivative, \(F'(x)\), which is:

• \(F'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)\)

This formula is key when the limits are functions of \(x\). In our specific example, we take the upper limit \(\sqrt{x}\) and the lower limit \(1\). Then, apply the formula to find the integral's rate of change, or its derivative. Remember, understanding this connection allows us to shift seamlessly between the worlds of integration and differentiation.

Application of Calculus

Calculus is not just a collection of rules; it serves as a powerful tool for solving complex problems, like determining the behavior of functions over time or space. In this context, the application of calculus goes beyond just differentiating or integrating.

In real-world contexts, you might deal with dynamic systems where understanding how things change with respect to one another is crucial. Being able to swap between integrals and their derivatives means you're ready to tackle challenges beyond simple computations.

• Predict stock movements based on historical data.
• Model physical phenomena like heat or motion.
• Optimize functions in engineering contexts, like minimizing material use while ensuring safety.
  
  It illustrates that mathematics is not just about answers, but about understanding the problems and dynamics themselves.

Definite Integral

A definite integral is an integral with limits. It's used to calculate the net area under a curve within a specified interval. In practical applications, it's the mathematical tool to determine quantities like area, total distance, or accumulated value between two points.

Consider the integral in our exercise: \(\int_{1}^{\sqrt{x}} \frac{z^2}{z^4 + 1} \, dz\). The limits \(1\) and \(\sqrt{x}\) make it definite, distinct from an indefinite integral which doesn't have limits.

• The lower limit (\(1\)) is fixed.
• The upper limit (\(\sqrt{x}\)) changes with \(x\).

With definite integrals, once computed over a range, you obtain a number. This tells you the total accumulation, a handy piece of information showing not just instantaneous change, but accumulated impact over the interval. It’s the sum of instantaneous effects over time or space.

Differentiation Rules

Differentiation is a central operation in calculus, used to find the rate of change of a function. There are several rules to master this process efficiently:

• Power Rule: For any term \(x^n\), the derivative is \(nx^{n-1}\).
• Product Rule: For two functions \(u(x)\) and \(v(x)\), the derivative is \(u'v + uv'\).
• Chain Rule: For a composite function \(f(g(x))\), the derivative is \(f'(g(x))g'(x)\).
  
  In our original problem, understanding these rules allows us to take the derivative of our integral and compute derivative expressions easily, especially when dealing with variable limits of integration. The chain rule plays a role when differentiating the upper limit function \(\sqrt{x}\) to find its derivative, \(\frac{1}{2\sqrt{x}}\). By familiarizing yourself with these rules, transforming integrals into derivatives becomes more intuitive and manageable.