/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Find the derivative of each func... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the derivative of each function. \(f(x)=\sqrt{3 x-2}\)

Short Answer

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

Step by step solution

01

Identify the Inner and Outer Functions

Here, the given function is \(f(x) = \sqrt{3x - 2}\), which can be expressed as \(f(x) = (\sqrt{x})((3x - 2))\), where the outer function \(u(x) = \sqrt{x}\) and the inner function \(v(x) = 3x - 2\).
02

Differentiate the Outer Function

We will differentiate the outer function \(u(x) = \sqrt{x}\) with respect to x. Recall that the power rule states that if \(u(x) = x^n\), then \(u'(x) = nx^{n-1}\). In our case, we have \(u(x) = x^{\frac{1}{2}}\). Applying the power rule gives: \(u'(x) = \frac{1}{2}x^{-\frac{1}{2}}\)
03

Differentiate the Inner Function

We will differentiate the inner function \(v(x) = 3x - 2\) with respect to x. Using the power rule again and considering that the derivative of a constant is zero, we get: \(v'(x) = 3\)
04

Apply the Chain Rule

The chain rule states that if a function f is the composition of two functions u and v, i.e., \(f(x) = u(v(x))\), then the derivative of f is given by: \(f'(x) = u'(v(x)) * v'(x)\) Now, substituting the derivative of the outer function u'(x), the inner function v(x), and the derivative of the inner function v'(x) from steps 2 and 3, we get: \(f'(x) = (\frac{1}{2}(3x - 2)^{-\frac{1}{2}}) * (3)\)
05

Simplify the Expression

Finally, we can simplify the expression of the derivative: \(f'(x) = \frac{3}{2}(3x - 2)^{-\frac{1}{2}}\) So, the derivative of the given function \(f(x) = \sqrt{3x - 2}\) is: \(f'(x) = \frac{3}{2\sqrt{3x - 2}}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Chain Rule
The chain rule is a fundamental technique in calculus used to determine the derivative of a composite function. When a function is composed of two or more functions, the chain rule allows us to differentiate the entire function by taking the product of the derivative of the outer function and the derivative of the inner function.

In the case of the given exercise, the function \( f(x) = \sqrt{3x - 2} \) can be seen as the composition of \( u(x) = \sqrt{x} \) and \( v(x) = 3x - 2 \). The chain rule lets us break down the complex function into simpler parts, differentiate those parts, and then piece it all back together to find the derivative of the original function. This methodical approach makes differentiation manageable even when the functions involved are more intricate.
Power Rule
The power rule is a basic rule in differentiation that simplifies the process of taking derivatives when dealing with powers of \( x \). When you have a function of the form \( u(x) = x^n \), the power rule tells us that the derivative of this function, \( u'(x) \), is \( nx^{n-1} \).

In our original exercise, we applied the power rule to both the inner and outer functions. For instance, the outer function \( u(x) = x^{\frac{1}{2}} \) has a power of \( \frac{1}{2} \), and thus, its derivative is \( \frac{1}{2}x^{-\frac{1}{2}} \) when we decrease the exponent by one. The power rule is a powerful tool that takes out the complexity in many types of differentiation problems.
Differentiation Steps
Differentiation often involves breaking down complex problems into a series of simpler steps. By following a structured approach, you ensure that you don't skip any important stages of the differentiation process. The steps typically start with identifying the function components, such as the inner and outer functions. Once recognized, the next steps include applying the relevant rules of differentiation, such as the power rule, to each component separately.

After finding the derivatives of individual pieces, you can combine them using the chain rule to get the complete derivative of the original function. If you follow these steps carefully, differentiating even the most complicated functions can become a systematic and error-free procedure, leading to a successful solution.
Inner and Outer Functions
Understanding inner and outer functions is vital when dealing with composite functions. The inner function is the function inside another function, while the outer function is the one that encompasses another function.

Identifying these is the first step in solving differentiation problems with the chain rule. In the given exercise, \( v(x) = 3x - 2 \) is the inner function because it is the part input into \( u(x) = \sqrt{x} \), the outer function. Recognizing the two allows us to apply the chain rule correctly. Focusing on the roles of inner and outer functions simplifies complex differentiation, leading students to deeper understanding and mastering this topic in calculus.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(f(x)=\frac{x}{x^{2}+1} .\) Find the point(s) on the graph of \(f\) where the tangent line is horizontal.

Find the derivative of each function. \(f(x)=2 x^{2}(3-4 x)^{4}\)

EFFECT OF HousING STARTS ON JoBS The president of a major housing construction firm claims that the number of construction jobs created is given by $$ N(x)=1.42 x $$ where \(x\) denotes the number of housing starts. Suppose the number of housing starts in the next \(t\) mo is expected to be $$ x(t)=\frac{7 t^{2}+140 t+700}{3 t^{2}+80 t+550} $$ million units/year. Find an expression that gives the rate at which the number of construction jobs will be created \(t\) mo from now. At what rate will construction jobs be created 1 yr from now?

The number of Americans aged 45 to 54 is approximately $$ \begin{aligned} N(t)=&-0.00233 t^{4}+0.00633 t^{3}-0.05417 t^{2} \\ &+1.3467 t+25 \end{aligned} $$ million people in year \(t\), with \(t=0\) corresponding to the beginning of 1990 . Compute \(N^{\prime}(10)\) and \(N^{\prime \prime}(10)\) and interpret your results. Source U.S. Census Bureau

TRAFFIC FLow Opened in the late \(1950 \mathrm{~s}\), the Central Artery in downtown Boston was designed to move 75,000 vehicles a day. The number of vehicles moved per day is approximated by the function $$ x=f(t)=6.25 t^{2}+19.75 t+74.75 \quad(0 \leq t \leq 5) $$ where \(x\) is measured in thousands and \(t\) in decades, with \(t=0\) corresponding to the beginning of \(1959 .\) Suppose the average speed of traffic flow in mph is given by $$ S=g(x)=-0.00075 x^{2}+67.5 \quad(75 \leq x \leq 350) $$ where \(x\) has the same meaning as before. What was the rate of change of the average speed of traffic flow at the beginning of \(1999 ?\) What was the average speed of traffic flow at that time? Hint: \(S=g[f(t)]\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.