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Chapter 4: Problem 84

In Exercises \(81-86,\) find \(F^{\prime}(x)\) . $$F(x)=\int_{2}^{x^{2}} \frac{1}{t^{3}} d t$$

Short Answer

Expert verified
So, the derivative \(F'(x)\) is given by \(F'(x) = -\frac{1}{x^{3}}\) for \(x>2\) and \(F'(x) = -\frac{1}{8}\) for \(x=2\).

Step by step solution

01

Apply the Fundamental Theorem of Calculus

By the fundamental theorem of calculus, given \(F(x)=\int_{2}^{x^{2}} \frac{1}{t^{3}} dt\), we have that \(F(x)=F(x^{2})-F(2)\) when we take F as the antiderivative of \(\frac{1}{t^{3}}\), i.e, \(-\frac{1}{2t^2}\). This results in \(F(x)=-\frac{1}{2x^{4}} - F(2)\).
02

Apply the Chain Rule

For the next part of the process, we apply the chain rule. The chain rule states that the derivative of \(f(g(x))\) is \(f'(g(x))g'(x)\). Hence \(F'(x) = -\frac{1}{2x^{4}} * 2x = -\frac{1}{x^{3}}\).
03

Compute F(2)

Now we know \(F'(x) = -\frac{1}{x^{3}}\) for x>2, since the lower limit of the original function was 2. Thus, for x=2, \(F'(x) = -\frac{1}{x^{3}} = -\frac{1}{2^{3}} = -\frac{1}{8}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Derivatives
Derivatives are a fundamental concept in calculus that measure how a function changes as its input changes. When you calculate a derivative, you are essentially finding the slope of the function at any given point. This slope represents the rate of change of the function's value with respect to changes in its input.

  • For example, the derivative of a position function with respect to time gives you velocity.
  • The derivative of a velocity function with respect to time gives you acceleration.
In the given exercise, we aim to find the derivative of a function defined by an integral, a task perfectly suited for the Fundamental Theorem of Calculus. The theorem allows us to connect differentiation and integration conveniently, making it possible to compute the derivative of integral functions.
Applying the Chain Rule
The chain rule is a crucial technique in calculus for finding the derivative of composite functions. It states that if a function is composed of two other functions, say \( f(g(x)) \), the derivative is given by \( f'(g(x)) \, g'(x) \).

This means you first find the derivative of the outer function evaluated at the inner function, and then multiply it by the derivative of the inner function.

  • In our exercise, the integral's upper limit is \( x^2 \), making use of the chain rule essential.
  • We take the derivative of \( \frac{1}{t^3} \) with respect to \( t \) and use the chain rule to evaluate it at \( x^2 \), then multiply by the derivative of \( x^2 \), which is \( 2x \).
  • The result simplifies to \( -\frac{1}{x^3} \).
Making use of the chain rule simplifies even the most complex-looking derivative problems, as seen in this exercise.
Concept of Definite Integrals
Definite integrals are a way to calculate the accumulation of a quantity, such as area under a curve, over a specified interval. Unlike indefinite integrals, which contain a constant of integration, definite integrals evaluate to a specific number.

  • They have limits of integration, such as \( \int_{2}^{x^2} \).
  • The value of a definite integral can be influenced by both the function being integrated and the specified limits.
In this exercise, we work with a definite integral, leveraging the limits \( 2 \) and \( x^2 \). The Fundamental Theorem of Calculus helps us find the derivative by expressing \( F(x) \) as the difference between the antiderivative evaluated at the upper and lower limits.

Thus, understanding definite integrals and their evaluation is key to solving such problems efficiently.

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