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Chapter 3: Problem 10

\(7 - 46\) Find the derivative of the function. $$f ( x ) = \left( 1 + x ^ { 4 } \right) ^ { 2 / 3 }$$

Short Answer

Expert verified
The derivative is \( f'(x) = \frac{8}{3} x^3 (1 + x^4)^{-1/3} \).

Step by step solution

01

Identify the function to differentiate

The function given is \( f(x) = (1 + x^4)^{2/3} \). We need to find its derivative, \( f'(x) \).
02

Apply the Chain Rule

We will use the Chain Rule, which states that if \( f(x) = g(h(x)) \), then \( f'(x) = g'(h(x)) \cdot h'(x) \). Here, \( g(u) = u^{2/3} \) and \( h(x) = 1 + x^4 \) with \( u = h(x) \).
03

Differentiate the outer function

Differentiate \( g(u) = u^{2/3} \) with respect to \( u \). The derivative is \( g'(u) = \frac{2}{3} u^{-1/3} \).
04

Differentiate the inner function

Differentiate \( h(x) = 1 + x^4 \) with respect to \( x \). The derivative is \( h'(x) = 4x^3 \).
05

Combine the results using the Chain Rule

Multiply the derivative of the outer function by the derivative of the inner function using the Chain Rule: \( f'(x) = g'(h(x)) \cdot h'(x) = \frac{2}{3} (1 + x^4)^{-1/3} \cdot 4x^3 \).
06

Simplify the expression

Simplify the expression for the derivative: \( f'(x) = \frac{8}{3} x^3 (1 + x^4)^{-1/3} \).

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Key Concepts

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

Chain Rule
The chain rule is a fundamental principle in calculus used to differentiate composite functions. Suppose you have a nested function like \( f(x) = (1 + x^4)^{2/3} \). Here, the composition is between the outer function \( g(u) = u^{2/3} \) and the inner function \( h(x) = 1 + x^4 \). The chain rule allows us to find the derivative by linking the rate of change of the outer function with the inner function.
  • The chain rule formula is: \( f'(x) = g'(h(x)) \cdot h'(x) \).
  • We differentiate the outer function \( g(u) \) with respect to \( u \), resulting in \( g'(u) = \frac{2}{3} u^{-1/3} \).
  • Also, we differentiate the inner function \( h(x) \) with respect to \( x \), giving \( h'(x) = 4x^3 \).
  • By combining these derivatives, you get the derivative of the composite function.
A step to remember when applying the chain rule is to carefully track which function you are differentiating at each step. This rule is incredibly useful when dealing with complex functions and understanding how they change.
Differentiation
Differentiation is the process of finding the derivative of a function. The derivative represents the rate of change or the slope of the function at a given point. For the function \( f(x) = (1 + x^4)^{2/3} \), differentiation involves several steps as shown in the solution.
  • First, identify the functions involved: here, \( g(u) = u^{2/3} \) and \( h(x) = 1 + x^4 \).
  • Next, use known derivative rules. For \( g(u) \), apply the power rule: \( g'(u) = \frac{2}{3}u^{-1/3} \).
  • For \( h(x) \), notice the simplicity of differentiating \( x^4 \): \( h'(x) = 4x^3 \).
  • Finally, combine the results using the chain rule to get the complete derivative.
Differentiation is key in calculus not only for functions like polynomials but also for wrapping complex computations into manageable parts, aiding in uncovering behavior of functions. This fundamental technique helps in various fields like physics, engineering, and economics.
Calculus
Calculus is a branch of mathematics that studies how things change. It is divided into differential calculus, which focuses on rates of change, and integral calculus, which deals with accumulation of quantities. The problem solving aspect, like the one handled here, is an example of differential calculus.The given function \( f(x) = (1 + x^4)^{2/3} \), illustrates how calculus allows us to calculate the behavior of functions through differentiation.
  • We begin by understanding the function's structure, recognizing opportunities to utilize the chain rule when dealing with composite functions.
  • Calculus provides systematic methods such as the power rule and product rule for direct computation of derivatives.
  • Utilizing these methods, one can find tangent lines, identify rates of change, and optimize functions in real-world applications.
  • Moreover, calculus injects precision into mathematical models, vital for scientific and economic analysis.
Calculus serves as the backbone of modern mathematics, providing tools to tackle intricate problems with a structured approach, thus furthering our understanding of dynamic systems.

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Most popular questions from this chapter

\(23-28\) Use a linear approximation (or differentials) to estimate the given number. $$ 1 / 1002 $$

Use differentials to estimate the amount of paint needed to apply a coat of paint 0.05 \(\mathrm{cm}\) thick to a hemispherical dome with diameter 50 \(\mathrm{m} .\)

One side of a right triangle is known to be 20 \(\mathrm{cm}\) long and the opposite angle is measured as \(30^{\circ},\) with a possible error of \(\pm 1^{\circ} .\) $$ \begin{array}{l}{\text { (a) Use differentials to estimate the error in computing the }} \\ {\text { length of the hypotenuse. }} \\ {\text { (b) What is the percentage error? }}\end{array} $$

On page 431 of Physics: Calculus, 2\(d\) ed, by Eugene Hecht (Pacific Grove, CA: Brooks/Cole, \(2000 ),\) in the course of deriving the formula \(\mathrm{T}=2 \pi \sqrt{\mathrm{L} / \mathrm{g}}\) for the period of a $$ \begin{array}{l}{\text { pendulum of length } L, \text { the author obtains the equation }} \\ {\mathrm{a}_{\mathrm{T}}=-g \sin \theta \text { for the tangential acceleration of the bob of the }}\end{array} $$ pendulum, He then says, "for small angles, the value of \(\theta\) in radians is very nearly the value of \(\sin \theta\) ; they differ by less than 2\(\%\) out to about \(20^{\circ} .\) $$ \begin{array}{c}{\text { (a) Verify the linear approximation at } 0 \text { for the sine function: }} \\ {\sin x=x}\end{array} $$ (b) Use a graphing device to determine the values of \(x\) for which sin \(x\) and \(x\) differ by less than 2\(\%\) . Then verify Hecht's statement by converting from radians to degrees.

Using principles from physics it can be shown that when a cable is hung between two poles, it takes the shape of a curve \(y=f(x)\) that satisfies the differential equation $$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\rho g}{\mathrm{T}} \sqrt{1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}}$$ where \(\rho\) is the linear density of the cable, \(g\) is the acceleration due to gravity, and T is the tension in the cable at its lowest point, and the coordinate system is chosen appropriately. Verify that the function $$y=f(x)=\frac{T}{\rho g} \cosh \left(\frac{\rho g x}{T}\right)$$ is a solution of this differential equation.
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