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Calculate the derivatives of the functions in Exercises 1-46. HINT [See Example 1.] \(f(x)=\left(2 x^{3}+x\right)^{-2}\)

Short Answer

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The short answer is: \(f'(x) = -2(6x^2 + 1)(2x^3 + x)^{-3}\)

Step by step solution

01

Find the derivatives of g(x) and h(x)

First, we need to find the derivatives of the functions \(g(x)\) and \(h(x)\). Apply the power rule to both functions to find their derivatives: \(g'(x) = \frac{d}{dx}(x^{-2}) = -2x^{-3}\) \(h'(x) = \frac{d}{dx}(2x^3 + x) = 6x^2 + 1\) Now that we have the derivatives, we can proceed to the chain rule.
02

Apply the chain rule

Next, we apply the chain rule to calculate the derivative of our function \(f(x)\): \(f'(x) = g'(h(x)) \cdot h'(x) = (-2[2x^3 + x]^{-3}) \cdot (6x^2 + 1)\)
03

Simplify the derivative

Now we just need to simplify our expression for the derivative. Begin by factoring out the term \((2x^3 + x)^{-3}\) from the expression: \(f'(x) = -2(6x^2 + 1)(2x^3 + x)^{-3}\) That's our final expression for the derivative of \(f(x)\): \(f'(x) = -2(6x^2 + 1)(2x^3 + x)^{-3}\)

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Key Concepts

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

Derivative Calculation
In calculus, derivative calculations are essential as they help us understand how functions change. When we calculate a derivative, we're finding how fast a function's value changes at any given point. This can show us trends like growth or decline and is vital in many fields like physics, engineering, and economics.
To start calculating a derivative, one of the most common methods used is applying derivative rules, among which the power rule and chain rule are particularly popular.
By carefully using these rules, you can determine the slope of a tangent line to the curve of a function at any point, which is precisely what the derivative represents. This fundamental tool assists in solving complex problems involving rates of change.
Power Rule
The power rule is a quick and easy way to find the derivative of functions in the form of \(x^n\), where \(n\) is any real number. Here's how it works: if you have a function \( f(x) = x^n \), the derivative is \( f'(x) = nx^{n-1} \).
This rule is an essential building block in calculus:
  • Reduces calculation time for polynomials.
  • Simple to apply across various functions.
  • Forms basis for more complex rules.
Using the power rule, you can quickly determine derivatives of polynomial expressions, like in our exercise.The power rule lets us find the derivative of \(x^{-2}\), resulting in \(-2x^{-3}\), and for a more complex term like \(2x^3 + x\), making its derivative \(6x^2 + 1\).
These easy calculations show the utility and efficiency of the power rule in simplifying derivative calculations.
Chain Rule
The chain rule is crucial for finding the derivative of composite functions; that is, when one function is nested within another.
This rule is like a chain reaction, hence its name! It involves deriving the outer function and then multiplying it by the derivative of the inner function.
Essentially, it follows this mathematical format: \( f'(x) = g'(h(x)) \times h'(x) \).
  • Must identify inner and outer functions.
  • Apply the derivative to each individually.
  • Multiply the derivatives together.
In our exercise, \(f(x) = (2x^3 + x)^{-2}\) is a composite function where \(h(x) = 2x^3 + x\) and the outer function \(g(x) = x^{-2}\). The chain rule helps us find \(f'(x)\) by multiplying the derivative of the nesting \(h(x)\) by \(-2(2x^3 + x)^{-3}\), yielding the complete derivative of the function.
Function Simplification
After deriving a function, it’s important to simplify the obtained derivative. Simplification makes the expression concise and manageable, which is helpful for further analysis or application.
Here are some steps:
  • Factor out common terms.
  • Reduce complex fractions or expressions.
  • Write in an easily interpretable format.
In the solution given for the exercise, we simplified by factoring out \((2x^3 + x)^{-3}\). This simplification made our derivative appear as \(-2(6x^2 + 1)(2x^3 + x)^{-3}\), much clearer and easier to use in downstream applications.
Being capable of simplifying derivatives efficiently not only makes calculations easier but also enhances your understanding of the mathematical relationships involved.

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Most popular questions from this chapter

Find the indicated derivatives in Exercises 47-54. In each case, the independent variable is a (unspecified) function oft. HINT [See Quick Example 2 on page 827.] \(s=\frac{1}{r^{3}}+r^{0.5} .\) Find \(\frac{d s}{d t}\).

The demand for the Cyberpunk II arcade video game is modeled by the logistic curve $$ q(t)=\frac{10,000}{1+0.5 e^{-0.4 t}} $$ where \(q(t)\) is the total number of units sold \(t\) months after its introduction. a. Use technology to estimate \(q^{\prime}(4)\). b. Assume that the manufacturers of Cyberpunk II sell each unit for \(\$ 800\). What is the company's marginal revenue \(d R / d q ?\) c. Use the chain rule to estimate the rate at which revenue is growing 4 months after the introduction of the video game.

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

Information Highway The amount of information transmitted each month in the early years of the Internet \((1988\) to 1994) can be modeled by the equation $$ q(t)=\frac{2 e^{0.69 t}}{3+1.5 e^{-0.4 t}} \quad(0 \leq t \leq 6) $$ where \(q\) is the amount of information transmitted each month in billions of data packets and \(t\) is the number of years since the start of \(1988 .^{33}\) a. Use technology to estimate \(q^{\prime}(2)\). b. Assume that it costs \(\$ 5\) to transmit a million packets of data. What is the marginal cost \(C^{\prime}(q)\) ? c. How fast was the cost increasing at the start of \(1990 ?\)

Calculate the derivatives of the functions in Exercises 1-46. HINT [See Example 1.] \(r(s)=\left(2 s+s^{0.5}\right)^{-1}\)

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