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Chapter 3: Problem 18

Find the derivatives of the functions in Exercises 17-28. \(z=\frac{4-3 x}{3 x^{2}+x}\)

Short Answer

Expert verified
The derivative is \( z' = \frac{9x^2 - 24x - 4}{(3x^2 + x)^2} \).

Step by step solution

01

Identify the Derivatives Rule to Use

Given the function \( z = \frac{4 - 3x}{3x^2 + x} \), the task is to find its derivative with respect to \( x \). Since the function is in the form of a quotient, we will use the quotient rule for differentiation.
02

Write Down the Quotient Rule Formula

The quotient rule is formulated as follows: if \( z = \frac{u}{v} \), then the derivative \( z' \) is given by \( z' = \frac{v \cdot u' - u \cdot v'}{v^2} \), where \( u = 4 - 3x \) and \( v = 3x^2 + x \).
03

Differentiate the Numerator and Denominator

Find \( u' \) and \( v' \):- For the numerator \( u = 4 - 3x \), the derivative is \( u' = -3 \).- For the denominator \( v = 3x^2 + x \), the derivative is \( v' = 6x + 1 \).
04

Substitute into the Quotient Rule Formula

Substitute \( u \), \( u' \), \( v \), and \( v' \) into the quotient rule formula. This gives:\[ z' = \frac{(3x^2 + x)(-3) - (4 - 3x)(6x + 1)}{(3x^2 + x)^2} \].
05

Simplify the Expression

Simplify the expression obtained from the quotient rule:\[ z' = \frac{(-9x^2 - 3x) - (24x + 4 - 18x^2 - 3x)}{(3x^2 + x)^2} \].Simplifying the terms gives:- Combine like terms in the numerator: \(-9x^2 - 3x - (24x + 4 - 18x^2 - 3x)\)- Simplify to get: \(9x^2 - 24x - 4\)So the derivative is:\[ z' = \frac{- 9x^2 - 3x - 24x - 4 + 18x^2 + 3x}{(3x^2 + x)^2} \].After simplification:\[ z' = \frac{9x^2 - 24x - 4}{(3x^2 + x)^2} \].

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Key Concepts

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

Quotient Rule
When encountering a function of the form \( \frac{u}{v} \), where both the numerator \( u \) and the denominator \( v \) are functions of \( x \), the quotient rule is your go-to method for finding derivatives. The quotient rule is stated as follows: if \( z = \frac{u}{v} \), the derivative \( z' \) is:
  • \( z' = \frac{v \cdot u' - u \cdot v'}{v^2} \)
Here, \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \), respectively. This rule is essential for differentiating functions that involve division. Be sure to apply the rule correctly by carefully plugging in your calculated derivatives into the formula.
Remember, the denominator in the derivative formula is \( v^2 \) which is squared. Misplacing signs or forgetting the \( v^2 \) can lead to errors, so caution is crucial.
Differentiation
Differentiation is the process of finding a derivative, which measures how a function changes as its input changes. It is a fundamental concept in calculus.
In our task, we are differentiating the function \( z = \frac{4 - 3x}{3x^2 + x} \). Differentiation gives us the rate of change of this function concerning the variable \( x \).
  • Identify which differentiation rule to apply based on the function form. For example, use the quotient rule when you have a quotient of two functions.
  • Apply basic differentiation rules such as the power rule and constant rule to find the derivatives of simpler sub-functions.
Differentiating lets us explore the behavior of functions, such as how steeply a curve rises or falls. This information is crucial in many real-world applications, like optimizing systems and predicting trends.
Numerator and Denominator Differentiation
Before using the quotient rule, it's important to differentiate both the numerator \( u \) and the denominator \( v \) of the function separately. This step is crucial because these derivatives, \( u' \) and \( v' \), become part of the final derivative calculation.
  • Numerator \( u = 4 - 3x \) has a derivative \( u' = -3 \).
  • Denominator \( v = 3x^2 + x \) is differentiated to get \( v' = 6x + 1 \).
After finding \( u' \) and \( v' \), substitute these derivatives into the quotient rule formula. This step ensures that the rate of change from both the numerator and denominator are accurately represented in the overall derivative. Make it a practice to always simplify if possible to provide a cleaner expression for the derivative. Being thorough with these steps can significantly reduce errors in complex differentiation problems.

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Most popular questions from this chapter

Use a CAS to perform the following steps for the functions \begin{equation} \begin{array}{l}{\text { a. Plot } y=f(x) \text { to see that function's global behavior. }} \\ {\text { b. Define the difference quotient } q \text { at a general point } x, \text { with }} \\ {\text { general step size } h .} \\\ {\text { c. Take the limit as } h \rightarrow 0 . \text { What formula does this give? }} \\ {\text { d. Substitute the value } x=x_{0} \text { and plot the function } y=f(x)} \\ \quad {\text { together with its tangent line at that point. }} \\ {\text { e. Substitute various values for } x \text { larger and smaller than } x_{0} \text { into }} \\ \quad {\text { the formula obtained in part (c). Do the numbers make sense }} \\ \quad {\text { with your picture? }} \\ {\text { f. Graph the formula obtained in part (c). What does it mean }} \\ \quad {\text { when its values are negative? Zero? Positive? Does this make }} \\ \quad {\text { sense with your plot from part (a)? Give reasons for your }} \\ \quad {\text { answer. }}\end{array} \end{equation} $$f(x)=x^{2} \cos x, \quad x_{0}=\pi / 4$$

If the original 24 \(\mathrm{m}\) edge length \(x\) of a cube decreases at the rate of 5 \(\mathrm{m} / \mathrm{min}\) , when \(x=3 \mathrm{m}\) at what rate does the cube's a. surface area change? b. volume change?

Changing voltage The voltage \(V\) (volts), current \(I\) (amperes) and resistance \(R\) (ohms) of an electric circuit like the one shown here are related by the equation \(V=I R .\) Suppose that \(V\) is increasing at the rate of 1 volt/sec while \(I\) is decreasing at the rate of 1\(/ 3\) amp/ sec. Let \(t\) denote time in seconds. a. What is the value of \(d V / d t ?\) b. What is the value of \(d I / d t ?\) c. What equation relates \(d R / d t\) to \(d V / d t\) and \(d I / d t\) ? d. Find the rate at which \(R\) is changing when \(V=12\) volts and \(I=2\) amps. Is \(R\) increasing, or decreasing?

Quadratic approximations $$ \begin{array}{l}{\text { a. Let } Q(x)=b_{0}+b_{1}(x-a)+b_{2}(x-a)^{2} \text { be a quadratic }} \\ {\text { approximation to } f(x) \text { at } x=a \text { with the properties: }}\end{array} $$ $$ \begin{aligned} \text { i) } Q(a) &=f(a) \\ \text { ii) } Q^{\prime}(a) &=f^{\prime}(a) \\ \text { iii) } Q^{\prime \prime}(a) &=f^{\prime \prime}(a) \end{aligned} $$ $$ \text{Determine the coefficients}b_{0}, b_{1}, \text { and } b_{2} $$ $$ \begin{array}{l}{\text { b. Find the quadratic approximation to } f(x)=1 /(1-x) \text { at }} \\ {\quad x=0 .} \\ {\text { c. } \operatorname{Graph} f(x)=1 /(1-x) \text { and its quadratic approximation at }} \\ {x=0 . \text { Then zoom in on the two graphs at the point }(0,1) \text { . }} \\ {\text { Comment on what you see. }}\end{array} $$ $$ \begin{array}{l}{\text { d. Find the quadratic approximation to } g(x)=1 / x \text { at } x=1} \\ {\text { Graph } g \text { and its quadratic approximation together. Comment }} \\ {\text { on what you see. }}\end{array} $$ $$ \begin{array}{l}{\text { e. Find the quadratic approximation to } h(x)=\sqrt{1+x} \text { at }} \\ {x=0 . \text { Graph } h \text { and its quadratic approximation together. }} \\ {\text { Comment on what you see. }} \\\ {\text { f. What are the linearizations of } f, g, \text { and } h \text { at the respective }} \\ {\text { points in parts (b), (d), and (e)? }}\end{array} $$

Particle acceleration A particle moves along the \(x\) -axis with velocity \(d x / d t=f(x) .\) Show that the particle's acceleration is \(f(x) f^{\prime}(x)\) .
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