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Find the derivatives of the functions in Exercises \(17-40 .\) $$z=\frac{4-3 x}{3 x^{2}+x}$$

Short Answer

Expert verified
The derivative is \( z'(x) = \frac{9x^2 - 27x - 4}{(3x^2 + x)^2} \).

Step by step solution

01

Understanding the Problem

We need to find the derivative of the function: \[ z = \frac{4 - 3x}{3x^2 + x} \] This is a rational function, so we'll employ the Quotient Rule for differentiation.
02

Recall the Quotient Rule

The Quotient Rule states that for a function \( f(x) = \frac{u(x)}{v(x)} \), the derivative \( f'(x) \) is given by: \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \] where \( u(x) \) and \( v(x) \) are functions of \( x \).
03

Identify the Components

Let \( u(x) = 4 - 3x \) and \( v(x) = 3x^2 + x \). We need to find their derivatives: \[ u'(x) = \frac{d}{dx}(4 - 3x) \] \[ v'(x) = \frac{d}{dx}(3x^2 + x) \]
04

Differentiate Numerator and Denominator

Differentiate the functions \( u(x) \) and \( v(x) \): \[ u'(x) = 0 - 3 = -3 \]\[ v'(x) = 6x + 1 \] Now we have the derivatives needed to apply the Quotient Rule.
05

Apply the Quotient Rule

Substitute \( u(x), u'(x), v(x), \) and \( v'(x) \) into the Quotient Rule formula:\[ z'(x) = \frac{(-3)(3x^2 + x) - (4 - 3x)(6x + 1)}{(3x^2 + x)^2} \]
06

Simplify the Expression

Calculate the expression step by step:\(-3(3x^2 + x) = -9x^2 - 3x \)\((4 - 3x)(6x + 1) = 24x + 4 - 18x^2 - 3x \)Thus, the numerator becomes:\(-9x^2 - 3x - (24x + 4 - 18x^2 - 3x) = 9x^2 - 27x - 4 \) The expression simplifies to:\[ z'(x) = \frac{9x^2 - 27x - 4}{(3x^2 + x)^2} \]
07

Final Answer

The derivative of the function \( z(x) = \frac{4-3x}{3x^2+x} \) is given by: \[ z'(x) = \frac{9x^2 - 27x - 4}{(3x^2 + x)^2} \]

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Key Concepts

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

Quotient Rule
The Quotient Rule is a handy tool in calculus for finding the derivative of a rational function, where one function is divided by another. It is essential when dealing with functions of the form \( f(x) = \frac{u(x)}{v(x)} \). To apply the Quotient Rule, follow these steps:
  • Identify the numerator as \( u(x) \) and the denominator as \( v(x) \).
  • Find the derivatives \( u'(x) \) and \( v'(x) \).
  • Plug these into the formula: \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]
This can be remembered as "low \( v(x) \), d-high \( u'(x) \), minus high \( u(x) \), d-low \( v'(x) \), all over the square of what's below."
It's crucial to apply every step correctly to avoid mistakes. Always simplify the final expression to ensure it's in its simplest form.
rational function
A rational function is a type of function that can be expressed as the ratio of two polynomials. It's written as \( z(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomial functions.These kinds of functions are ubiquitous in mathematics and appear frequently in problems involving derivatives, limits, and integrals. Here are key points to remember about rational functions:
  • The denominator \( Q(x) \) must not be zero because division by zero is undefined.
  • They can exhibit asymptotic behavior as they approach values that make the denominator zero.
  • They are usually more complex to differentiate due to their structure, which is why the Quotient Rule is used.
Understanding the components of a rational function helps simplify the process of differentiation, as it allows for the correct identification and targeting of each polynomial part.
differentiation
Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes at any given point. It is central to understanding the behavior of functions and is used to solve various real-world problems involving rates and changes. For rational functions, differentiation can initially seem complex due to the interaction of the polynomials in the numerator and denominator. Here are some key elements of differentiation:
  • The derivative of a polynomial is found by applying the power rule: lower the exponent by one and multiply by the original exponent.
  • The differentiation of multiple terms must consider each term separately before combining them.
  • Techniques such as the Quotient Rule are essential here to deal with more complex structures like a rational function.
Differentiation provides the slope of a function at a point, giving insight into its increasing or decreasing behavior. By practice and repetition with such rules, mastering these steps becomes manageable and highly beneficial for deeper mathematical understanding.

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Most popular questions from this chapter

The effect of flight maneuvers on the heart The amount of work done by the heart's main pumping chamber, the left ventricle, is given by the equation $$ W=P V+\frac{V \delta v^{2}}{2 g} $$ where \(W\) is the work per unit time, \(P\) is the average blood pressure, \(V\) is the volume of blood pumped out during the unit of time, \(\delta\) ("delta") is the weight density of the blood, \(v\) is the average velocity of the exiting blood, and \(g\) is the acceleration of gravity. When \(P, V, \delta,\) and \(v\) remain constant, \(W\) becomes a function of \(g\), and the equation takes the simplified form $$ W=a+\frac{b}{g}(a, b \text { constant }) $$ As a member of NASA's medical team, you want to know how sensitive \(W\) is to apparent changes in \(g\) caused by flight maneuvers, and this depends on the initial value of \(g\). As part of your investigation, you decide to compare the effect on \(W\) of a given change \(d g\) on the moon, where \(g=5.2 \mathrm{ft} / \mathrm{sec}^{2}\), with the effect the same change \(d_{8}\) would have on Earth, where \(g=32 \mathrm{ft} / \mathrm{sec}^{2}\). Use the simplified equation above to find the ratio of \(d W_{\text {moon }}\) to \(d W_{\text {Barth }}\)

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Use a CAS to perform the following steps. a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P\) c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P .\) Then plot the implicit curve and tangent line together on a single graph. $$x \sqrt{1+2 y}+y=x^{2}, \quad P(1,0)$$

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