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For the following exercises, find \(f^{\prime}(x)\) for each function. $$f(x)=x^{2}\left(\frac{2}{x^{2}}+\frac{5}{x^{3}}\right)$$

Short Answer

Expert verified
The derivative is \(f'(x) = -\frac{5}{x^2}.\)

Step by step solution

01

Distribute Terms Inside the Parentheses

First, distribute the term \(x^2\) into each term inside the parentheses. This gives:\[f(x) = x^2 \cdot \frac{2}{x^2} + x^2 \cdot \frac{5}{x^3}.\]
02

Simplify the Expression

Simplify each term by canceling out the common factors. - \(x^2 \cdot \frac{2}{x^2}\) simplifies to \(2\) because the \(x^2\) terms cancel out.- \(x^2 \cdot \frac{5}{x^3}\) simplifies to \(\frac{5}{x}\) because \(x^2/x^3 = 1/x\).Thus, the simplified expression is:\[f(x) = 2 + \frac{5}{x}.\]
03

Convert the Expression for Differentiation

Rewrite the expression for convenience in differentiation:\[f(x) = 2 + 5x^{-1}.\]
04

Differentiate Each Term

Differentiate each term of \(f(x) = 2 + 5x^{-1}\).- The derivative of a constant \(2\) is \(0\).- The derivative of \(5x^{-1}\) is found using the power rule: \(-5x^{-2}\).So, \(f'(x) = 0 - 5x^{-2}.\)
05

Simplify the Derivative

Combine the results to get the final derivative:\[f'(x) = -\frac{5}{x^2}.\]

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus. It involves finding the derivative of a function, which measures how the function's output changes as its input changes. In simpler terms, the derivative tells you the rate at which one quantity changes with respect to another. In this exercise, we need to find the derivative of the given function, which will tell us how the function behaves as the input, or the variable, changes.

When you differentiate a function, you're essentially calculating the slope of the tangent line at any given point along the curve represented by that function. This slope provides a great deal of insight into the function's behavior, such as whether it's increasing or decreasing, and by how much, in different sections of its graph.
Power Rule
The power rule is a quick and efficient method for finding the derivative of a term that is in the form of a power. Specifically, if you have a term like \(x^n\), the power rule tells you that the derivative is \(nx^{n-1}\). It's a handy shortcut for differentiation, especially when dealing with simple polynomials.

In our exercise, the term \(5x^{-1}\) was differentiated using the power rule. To apply it, we multiplied the power by the coefficient, resulting in \(-5\), and then decreased the exponent by one to get \(-2\). Therefore, the derivative of \(5x^{-1}\) becomes \(-5x^{-2}\). Mastering the power rule can greatly simplify calculus problems by reducing complex differentiation to a series of simple steps.
Simplification
Simplification involves reducing an expression to its simplest form. This practice is crucial because it makes mathematical operations more manageable and clearer to work with. In the given exercise, simplification was an essential step before applying differentiation.

Initially, we had the expression \(x^2\left( \frac{2}{x^2} + \frac{5}{x^3} \right)\), which looks complex. By distributing \(x^2\) and then simplifying each term—specifically \(x^2 \cdot \frac{2}{x^2}\) to \(2\), and \(x^2 \cdot \frac{5}{x^3}\) to \(\frac{5}{x}\)—the expression became much simpler: \(2 + 5x^{-1}\). This simplification step recasts the problem into a much easier form for differentiation.
Function
A function is a relation that uniquely associates an input with a corresponding output. In mathematics, it is a rule that assigns to each element in a set exactly one element in another set. Functions are often represented as equations, like the one in this exercise: \(f(x) = x^2\left( \frac{2}{x^2} + \frac{5}{x^3} \right)\).

Understanding the properties and behavior of a function is key when working with calculus. Knowing how a function changes (the rate of change) involves studying its derivative. In our exercise, we analyzed and simplified the function, and then derived it, to understand how it changes with respect to \(x\). This understanding is foundational for solving more complex problems in calculus and applied mathematics.

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Most popular questions from this chapter

For the following exercises, find \(f^{\prime}(x)\) for each function. $$f(x)=2^{x} \cdot \log _{3} 7^{x^{2}-4}$$

For the following exercises, use the given values to find \(\left(f^{-1}\right)^{\prime}(a).\) $$f(1)=-3, f^{\prime}(1)=10, a=-3$$

[T] The position in feet of a race car along a straight track after \(t\) seconds is modeled by the function $$ s(t)=8 t^{2}-\frac{1}{16} t^{3} $$ a. Find the average velocity of the vehicle over the following time intervals to four decimal places: $$ \begin{array}{l}{\text { i. }[4,4.1]} \\ {\text { ii. }[4,4.01]} \\ {\text { iii. }[4,4.001]} \\ {\text { iv. }[4,4,0001]}\end{array} $$ b. Use a. to draw a conclusion about the instantaneous velocity of the vehicle at \(t=4\) seconds.

[T] An isotope of the element erbium has a half-life of approximately 12 hours. Initially there are 9 grams of the isotope present. a. Write the exponential function that relates the amount of substance remaining as a function of t, measured in hours. b. Use a. to determine the rate at which the substance is decaying in t hours. c. Use b. to determine the rate of decay at t = 4 hours.

[T] The relative rate of change of a differentiable function \(y=f(x)\) is given by \(\frac{100 \cdot f^{\prime}(x)}{f(x)} \% .\) One model for population growth is a Gompertz growth function, given by \(P(x)=a e^{-b \cdot e^{-c x}}\) where \(a, b, \quad\) and \(c\) are constants. a. Find the relative rate of change formula for the generic Gompertz function. b. Use a. to find the relative rate of change of a population in x = 20 months when a = 204, b = 0.0198, and c = 0.15. c. Briefly interpret what the result of b. means.

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