Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Finding a Derivative In Exercises \(7-34,\) find the derivative of the function. $$ f(x)=\sqrt[3]{12 x-5} $$

Short Answer

Expert verified
The derivative of the function \(f(x) = \sqrt[3]{12x - 5}\) is \(f'(x) = \frac{4}{(12x - 5)^{2/3}}\).

Step by step solution

01

Rewrite the Function

Rewrite the function \(f(x)\) in the form that can be more easily differentiated. In this case, it's better to write it in its power form: \(f(x) = (12x - 5)^{1/3}\).
02

Apply the Chain Rule

Apply the chain rule, which states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. Here the outer function is \(u^{1/3}\) and the inner function is \(u=12x - 5\). So we have: \(f'(x) = \frac{1}{3}(12x - 5)^{-2/3} \cdot (12)\).
03

Simplify the Result

Now let's simplify the given expression: \(f'(x) = \frac{12}{3}(12x - 5)^{-2/3} = 4(12x - 5)^{-2/3}\). This expression simplifies to: \(f'(x) = \frac{4}{(12x - 5)^{2/3}}\).

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 Vaia!

Key Concepts

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

Chain Rule
The chain rule is a fundamental tool in calculus used for finding the derivative of a composite function. A composite function is a function made up of two functions. For example, if you have a function inside another function, you'll need to use the chain rule to differentiate it properly.
The chain rule states that the derivative of a composite function is the product of the derivative of the outer function and the derivative of the inner function. In notation, if you have a function like \(f(g(x))\), the derivative is given by \(f'(g(x)) \, \cdot\, g'(x)\). This means you first find the derivative of the outer function while keeping the inner function the same, and then multiply it by the derivative of the inner function.
Applying the chain rule helps when functions are layered and ensures the transition between functions is accounted for in the calculation of the derivative.
Power Rule
The power rule is one of the simplest and most frequently used rules in differentiation. It is used when dealing with functions of the form \(x^n\). According to the power rule, if you have \(f(x) = x^n\), then the derivative is \(f'(x) = n \cdot x^{n-1}\).
This rule works by lowering the power by one and multiplying by the original power, simplifying the process of differentiation significantly.
  • For example, if you have \(f(x) = x^3\), then the derivative is \(f'(x) = 3x^2\).
  • The power rule is applicable for both positive and negative exponents.
  • It's very handy for quickly finding derivatives of polynomials and power-based functions.
In our original equation \(f(x) = (12x - 5)^{1/3}\), the power rule is used in combination with the chain rule to find the derivative.
Function Differentiation
Function differentiation is the process of finding the derivative of a function. The derivative represents the rate of change of the function with respect to a variable. In simpler terms, it tells you how much the function's output changes when you change the input slightly.
  • Differentiation helps in understanding how a function behaves, such as whether it's increasing or decreasing.
  • Derivatives have numerous applications, from physics (to find velocity and acceleration) to economics (for cost and revenue analysis).
  • It involves the application of rules like the sum rule, product rule, and others, depending on the structure of the function.
Finding the derivative involves algebraic manipulations and applying different rules of calculus. In this exercise, differentiation involves rewriting the function to a power form, using the power and chain rules to find the derivative.
Composite Function
A composite function is created when one function is applied to the result of another function. In the context of calculus, understanding composite functions is crucial, especially when it comes to differentiation.
A composite function is expressed as \(f(g(x))\), meaning function \(f\) is applied to the result of \(g(x)\). In our case, the function \(f(x) = \sqrt[3]{12x-5}\) or equivalently \(f(x) = (12x-5)^{1/3}\) is composite because it applies the power function to the linear expression \(12x-5\).
  • The chain rule is necessary for differentiating these kinds of functions, as it involves multiple layers.
  • Composites are commonly encountered in real-world problems where processes are dependent on other processes.
Understanding how to work with composite functions is essential for effective calculus problem-solving. It allows you to break down complex tasks into simpler, manageable parts.

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

Modeling Data The table shows the health care expenditures \(h\) (in billions of dollars) in the United States and the population \(p\) (in millions) of the United States for the years 2004 through 2009 . The year is represented by \(t,\) with \(t=4\) corresponding to 2004 . (Source: U.S. Centers for Medicare \& Medicaid Services and U.S. Census Bureau) $$ \begin{array}{|c|c|c|c|c|c|}\hline \text { Year, } & {4} & {5} & {6} & {7} & {8} & {9} \\ \hline h & {1773} & {1890} & {2017} & {2135} & {2234} & {2330} \\\ \hline p & {293} & {296} & {299} & {302} & {305} & {307} \\\ \hline\end{array} $$ (a) Use a graphing utility to find linear models for the health care expenditures \(h(t)\) and the population \(p(t) .\) (b) Use a graphing utility to graph each model found in part (a). (c) Find \(A=h(t) / p(t),\) then graph \(A\) using a graphing utility. What does this function represent? (d) Find and interpret \(A^{\prime}(t)\) in the context of these data.

Linear and Quadratic Approximations The linear and quadratic approximations of a function \(f\) at \(x=a\) are $$\begin{array}{l}{P_{1}(x)=f^{\prime}(a)(x-a)+f(a) \text { and }} \\\ {P_{2}(x)=\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}+f^{\prime}(a)(x-a)+f(a)}\end{array}$$ In Exercises 123 and \(124,\) (a) find the specified linear and quadratic approximations of \(f,(b)\) use a graphing utility to graph \(f\) and the approximations, (c) determine whether \(P_{1}\) or \(P_{2}\) is the better approximation, and (d) state how the accuracy changes as you move farther from \(x=a\) . $$ f(x)=\sec x ; \quad a=\frac{\pi}{6} $$

Differential Equations In Exercises \(125-128\) , verify that the function satisfies the differential equation. $$ \text{Function} \quad \text{Differential Equation} $$ $$ y=3 \cos x+\sin x \quad y^{\prime \prime}+y=0 $$

Writing Use a graphing utility to graph the two functions \(f(x)=x^{2}+1\) and \(g(x)=|x|+1\) in the same viewing window. Use the zoom and trace features to analyze the graphs near the point \((0,1)\) . What do you observe? Which function is differentiable at this point? Write a short paragraph describing the geometric significance of differentiability at a point.

Area The radius \(r\) of a circle is increasing at a rate of 4 centimeters per minute. Find the rates of change of the area when (a) \(r=8\) centimeters and (b) \(r=32\) centimeters.

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.

Sign-up for free