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Chapter 3: Problem 31

Find the derivatives of the functions in Exercises \(23-50\). $$y=\frac{1}{18}(3 x-2)^{6}+\left(4-\frac{1}{2 x^{2}}\right)^{-1}$$

Short Answer

Expert verified
The derivative is \((3x-2)^5 - \frac{1}{x^3} \left(4 - \frac{1}{2x^2}\right)^{-2}\)."

Step by step solution

01

Identify the Components of the Function

The given function is\[ y = \frac{1}{18}(3x - 2)^6 + \left(4 - \frac{1}{2x^2}\right)^{-1} \]We need to find the derivative of this function, which consists of two separate terms: a power function term and a reciprocal function term.
02

Differentiate the Power Function Term

The first term is \( \frac{1}{18} (3x - 2)^6 \). Use the chain rule to differentiate it:- First identify \( u = 3x - 2 \) and thus \( u^6 = (3x - 2)^6 \).- Differentiate using \( \frac{d}{dx} [u^6] = 6u^5 \cdot \frac{du}{dx} \).- Since \( \frac{du}{dx} = 3 \), the derivative becomes:\[ \frac{1}{18} \cdot 6(3x - 2)^5 \cdot 3 = \frac{18}{18}(3x - 2)^5 = (3x - 2)^5 \]
03

Differentiate the Reciprocal Function Term

The second term is \( \left( 4 - \frac{1}{2x^2} \right)^{-1} \). Use the chain rule and quotient differentiation:- Let \( v = 4 - \frac{1}{2x^2} \) so the term is \( v^{-1} \).- Derivative of \( v^{-1} \) is \( -v^{-2} \cdot \frac{dv}{dx} \).- First, find \( \frac{dv}{dx} = \frac{d}{dx} \left( 4 - \frac{1}{2x^2} \right) = \frac{2}{2x^3} = \frac{1}{x^3} \).- Hence, derivative becomes:\[ -\left( 4 - \frac{1}{2x^2} \right)^{-2} \cdot \frac{1}{x^3} \]
04

Combine the Derivatives

Add the derivatives from both terms:\[ y' = (3x - 2)^5 - \frac{1}{x^3} \left( 4 - \frac{1}{2x^2} \right)^{-2} \]This represents the derivative of the entire function.

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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 powerful technique in calculus for finding the derivative of composite functions. A composite function is where one function, say \( g(x) \), is nested inside another function, \( f(u) \). It helps you differentiate complex functions by breaking them down into simpler parts.
To use the chain rule, follow these steps:
  • Identify the inner function \( u(x) \) and the outer function \( f(u) \).
  • Differentiate the outer function with respect to its variable \( u \), which gives \( f'(u) \).
  • Differentiate the inner function \( u(x) \) with respect to \( x \), which gives \( u'(x) \).
  • Combine these using the formula \( \frac{d}{dx}[f(u(x))]=f'(u(x)) \cdot u'(x) \).
This method is particularly useful when dealing with functions like power functions or other polynomials nested inside a larger framework, as seen in the given function.
Power Functions
Power functions have the form \( y = x^n \), where \( n \) is any real number. When differentiating power functions, the power rule is typically used. It states that the derivative of \( x^n \) is \( nx^{n-1} \). However, when combined with the chain rule, it can help tackle more complex expressions like \((3x - 2)^6\).
For example:
  • Set \( u = 3x - 2 \), and apply the power rule\((u^6)' = 6u^5 \).
  • Find the derivative of \( u \), \( u'(x) = 3 \).
  • Use the chain rule to get the full derivative of \( (3x - 2)^6 \) as \( 6(3x - 2)^5 \cdot 3 \), simplifying to \( (3x - 2)^5 \).
Thus, combining the chain rule with power functions is a crucial skill for calculus problems involving derivatives.
Reciprocal Functions
Reciprocal functions are those that can be expressed in the form \( \frac{1}{f(x)} \), where \( f(x) \) is a function of \( x \). Differentiating reciprocal functions can also require the chain rule along with the quotient rule when necessary.
In this scenario, you see terms like \( \left(4 - \frac{1}{2x^2}\right)^{-1} \).
Steps to differentiate reciprocal functions:
  • Express the function as \( v^{-1} \) where \( v = 4 - \frac{1}{2x^2} \).
  • Differentiate \( v^{-1} \) by using the chain rule where \( (v^{-1})' = -v^{-2} \).
  • Compute \( \frac{dv}{dx} \) and apply it to the derivative of \( v^{-1} \).
This results in \( -\frac{1}{x^3}\left(4 - \frac{1}{2x^2}\right)^{-2} \) as the derivative, showing how reciprocal functions are approached with calculus techniques.

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Most popular questions from this chapter

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I .\) Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\). c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying $$|x-a|<\delta \Rightarrow|f(x)-L(x)|<\epsilon$$ for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=\frac{x-1}{4 x^{2}+1},\left[-\frac{3}{4}, 1\right], \quad a=\frac{1}{2}$$

Find the derivative of \(y\) with respect to the given independent variable. $$y=\log _{25} e^{x}-\log _{5} \sqrt{x}$$

You will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS: a. Plot the function \(y=f(x)\) together with its derivative over the given interval. Explain why you know that \(f\) is one-to-one over the interval. b. Solve the equation \(y=f(x)\) for \(x\) as a function of \(y,\) and name the resulting inverse function \(g\). c. Find the equation for the tangent line to \(f\) at the specified point \(\left(x_{0}, f\left(x_{0}\right)\right)\) d. Find the equation for the tangent line to \(g\) at the point \(\left(f\left(x_{0}\right), x_{0}\right)\) located symmetrically across the \(45^{\circ}\) line \(y=x\) (which is the graph of the identity function). Use Theorem 3 to find the slope of this tangent line. e. Plot the functions \(f\) and \(g\), the identity, the two tangent lines, and the line segment joining the points \(\left(x_{0}, f\left(x_{0}\right)\right)\) and \(\left(f\left(x_{0}\right), x_{0}\right)\) Discuss the symmetries you see across the main diagonal. $$y=x^{3}-3 x^{2}-1, \quad 2 \leq x \leq 5, \quad x_{0}=\frac{27}{10}$$

A highway patrol plane flies 3 mi above a level, straight road at a steady \(120 \mathrm{mi} / \mathrm{h}\). The pilot sees an oncoming car and with radar determines that at the instant the line-of-sight distance from plane to car is 5 mi, the line-of-sight distance is decreasing at the rate of \(160 \mathrm{mi} / \mathrm{h}\). Find the car's speed along the highway.

At what rate is the angle between a clock's minute and hour hands changing at 4 o'clock in the afternoon?
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