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Calculate the derivatives of the functions in Exercises 1-46. HINT [See Example 1.] \(f(x)=(2-x)^{-2}\)

Short Answer

Expert verified
The derivative of the function \(f(x) = (2-x)^{-2}\) is \(f'(x) = 2(2-x)^{-3}\).

Step by step solution

01

Identify the inner function

First, let's identify the inner function, which we'll call \(u\). In this case, we have \(u(x) = 2 - x\). Now, we can rewrite our main function as: \(f(x) = (u(x))^{-2}\)
02

Compute derivative of the inner function

Next, we will compute the derivative of the inner function \(u(x) = 2 - x\). Using the power rule, we have: \(u'(x) = -1\)
03

Compute derivative of the main function with respect to the inner function

Now, we will find the derivative of the main function with respect to the inner function. That is, we will find the derivative of \(f(u) = u^{-2}\). Using the power rule, we have: \(f'(u) = -2u^{-3}\)
04

Apply the chain rule

Finally, we will apply the chain rule to find the derivative of the main function with respect to x. The chain rule states that: \(f'(x) = f'(u) \times u'(x)\) Plugging in the values of \(f'(u)\) and \(u'(x)\) from steps 2 and 3, we get: \(f'(x) = (-2u^{-3}) \times (-1)\) Now, don't forget to replace \(u\) back with its original expression in terms of x: \(f'(x) = (-2(2-x)^{-3}) \times (-1)\)
05

Simplify the expression

Now, let's simplify our expression: \(f'(x) = 2(2-x)^{-3}\) And that is our final answer. The derivative of the given function is: \(f'(x) = 2(2-x)^{-3}\)

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Key Concepts

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

Chain Rule
The chain rule is a fundamental principle in calculus that allows us to compute the derivative of composite functions—functions made up of other functions. When faced with a function like \(f(g(x))\), where \((g(x))\) is nested inside another function \(f\), it might seem a bit daunting to find its derivative directly. However, the chain rule simplifies this task by breaking it down into manageable steps.

The rule states that the derivative of \(f(g(x))\) with respect to \((x)\) is the product of the derivative of \(f\) with respect to \(g(x)\) and the derivative of \(g(x)\) with respect to \((x)\), or simply \(f'(x) = f'(g(x)) \times g'(x)\). This technique was utilized in the original exercise to find the derivative of the function \(f(x)=(2-x)^{-2}\), yielding a clear and methodical approach to what might initially appear as a complex problem.
Power Rule
When calculating derivatives, the power rule is perhaps the most frequently applied tool in your calculus toolkit. This rule makes finding derivatives of polynomials as straightforward as it gets. The power rule states that if \(f(x) = x^n\), then its derivative, \(f'(x)\), is \(nx^{n-1}\).

In our example, we applied the power rule twice: first to find the derivative of the inner function \(u(x) = 2 - x\), which is a linear function, where the application is simply a consequence of understanding that any constant derivative equals zero and the derivative of \(x\) is 1; and second to find the derivative of \(u^{-2}\), a power of the inner function. By following the power rule, we were able to swiftly move through these steps and gain a solid understanding of how derivatives transform when applying exponents.
Derivative Calculation
Calculating derivatives lies at the heart of calculus and is essential for understanding how quantities change. Derivative calculation involves understanding various rules like the chain rule and power rule—as seen in the exercise solution—as well as others like the product rule and quotient rule. The ability to calculate derivatives accurately allows one to analyze motion, optimize processes, and model real-life situations mathematically.

This process often involves identifying the types of functions involved, applying the appropriate rules, and simplifying the result to make it as clear and understandable as possible. As exemplified by the step-by-step solution, each stage of derivative calculation is critical, and skipping steps can lead to errors and misunderstandings.
Calculus Exercises
Calculus exercises like the one provided in the textbook solution are designed to reinforce the understanding of derivative calculation principles. Implementing these principles repeatedly in a variety of exercises helps build competence and confidence in solving calculus problems.

These exercises often start with simpler tasks and progressively introduce more complex ones, which may combine multiple rules and functions. Keeping organized and methodical, as shown in the textbook solution, is essential for success in calculus exercises. Students are encouraged to practice regularly, using the step-by-step format to deeply understand each concept and to avoid common pitfalls in calculus computations.

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Most popular questions from this chapter

The percentage y (of total personal consumption) an individual spends on food is approximately y = 35x?0.25 percentage points (6.5 ? x ? 17.5) where x is the percentage the individual spends on education.28 An individual finds that she is spending x = 7 + 0.2t percent of her personal consumption on education, where t is time in months since January 1. Use direct substitution to express the percentage y as a function of time t (do not simplify the expression) and then use the chain rule to estimate how fast the percentage she spends on food is changing on November 1. Be sure to specify the units.

Find the equation of the line tangent to the graph of the given function at the point with the indicated \(x\) -coordinate. \(f(x)=\frac{x}{x^{2}+1} ; x=1\)

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(based on a question from the GRE Economics Test) Which of the following statements about average cost and marginal cost is correct? (A) If average cost is increasing, marginal cost must be increasing. (B) If average cost is increasing, marginal cost must be decreasing. (C) If average cost is increasing, marginal cost must be more than average cost. (D) If marginal cost is increasing, average cost must be increasing. (E) If marginal cost is increasing, average cost must be larger than marginal cost.

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