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Chapter 11: Problem 18

Calculate \(\frac{d y}{d x}\) in Exercises 13-20. Simplify your answer. HINT [See Example 5.] \(y=(4 x-1)^{2}\)

Short Answer

Expert verified
The simplified derivative of \(y=(4x-1)^{2}\) is \(\frac{dy}{dx} = 8(4x-1)\).

Step by step solution

01

Identify the inner and outer functions

In this problem, we have \(y = (4x-1)^2\). The inner function is \(u = 4x - 1\), and the outer function is \(y = u^2\).
02

Apply the chain rule

The chain rule states that \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\). We need to find the derivatives of the inner function and outer function separately.
03

Find the derivative of the inner function

Find \(\frac{du}{dx}\) for the inner function \(u = 4x - 1\). Using the power rule, we have: \(\frac{du}{dx} = \frac{d}{dx}(4x - 1) = 4\)
04

Find the derivative of the outer function

Find \(\frac{dy}{du}\) for the outer function \(y = u^2\). Using the power rule, we have: \(\frac{dy}{du} = \frac{d}{du}(u^2) = 2u\)
05

Substitute back the inner function

Now, substitute back the inner function \(u = 4x - 1\) into the outer function's derivative: \(\frac{dy}{du} = 2u = 2(4x - 1)\)
06

Multiply the derivatives

Multiply the derivatives \(\frac{dy}{du}\) and \(\frac{du}{dx}\) using the chain rule: \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2(4x - 1) \cdot 4\)
07

Simplify the answer

Simplify the derivative: \(\frac{dy}{dx} = 2(4x - 1) \cdot 4 = 8(4x - 1)\) So, the derivative of \(y=(4x-1)^{2}\) is \(\frac{dy}{dx} = 8(4x-1)\).

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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 pivotal concept in calculus. It helps us find the derivative of composite functions - functions made up of an "inner" and an "outer" function. When we have something like
  • \(y = (4x-1)^2\),
the chain rule demands that you take the derivative of each layer in sequence. The formula for the chain rule is
  • \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).
You'll calculate the derivative of the outer function with respect to the inner one, then multiply that by the derivative of the inner function with respect to \(x\). This allows us to "chain" our derivatives together for accurate results.
Understanding how to apply the chain rule effectively opens the door to finding derivatives for a wide array of complex functions!
Inner and Outer Functions
In differentiating composite functions, identifying inner and outer functions is your first step.
  • The **inner function** is the part of the function on which another function is performed.
For instance, with
  • \(y = (4x-1)^2\),
the expression "\(4x - 1\)" is the inner function, denoted as \(u = 4x - 1\).
  • The **outer function** acts on the outcome of the inner function.
In this case, "\(u^2\)" is the outer function. Figuring out which part is inner and which is outer is crucial before applying the chain rule. Once you have each part set, you can use this structure to find derivatives confidently and systematically!
Power Rule
The power rule is a simple yet powerful tool in differentiation. It is used when you have a function in the form
  • \(x^n\),
where \(n\) can be any real number. The power rule states that:
  • \( \frac{d}{dx}(x^n) = nx^{n-1} \).
When applying the chain rule, we often use the power rule to differentiate both the inner and the outer functions. For
  • \( y = (4x-1)^2 \),
the outer function is \(u^2\), so the derivative is
  • \(2u^{2-1} = 2u\).
For the inner function \(4x-1\), the derivative is \(4\) because it is a linear function and thus adjusts directly through the constant. Combining these derivatives using the chain rule allows us to find \( \frac{dy}{dx} \) seamlessly.

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Most popular questions from this chapter

Formulate a simple procedure for deciding whether to apply first the chain rule, the product rule, or the quotient rule when finding the derivative of a function.

The percentage y (of total personal consumption) an individual spends on food is approximately y = 35x?0.25 percentage points (6.5 ? x ? 17.5) where x is the percentage the individual spends on education.28 An individual finds that she is spending x = 7 + 0.2t percent of her personal consumption on education, where t is time in months since January 1. Use direct substitution to express the percentage y as a function of time t (do not simplify the expression) and then use the chain rule to estimate how fast the percentage she spends on food is changing on November 1. Be sure to specify the units.

Compute the indicated derivative using the chain rule. HINT [See Quick Examples on page 828.] \(x=1-t / 2, y=4 t-1 ; \frac{d y}{d x}\)

Calculate the derivatives of the functions in Exercises 1-46. HINT [See Example 1.] \(f(x)=(x-1)^{-1}\)

The following graph shows the approximate value of home prices and existing home sales in 2006鈥�2010 as a percentage change from 2003, together with quadratic approximations The quadratic approximations are given by $$ \text { Home Prices: } \quad P(t)=t^{2}-10 t+41 \quad(0 \leq t \leq 4) $$ Existing Home Sales: \(\quad S(t)=1.5 t^{2}-11 t \quad(0 \leq t \leq 4)\) where \(t\) is time in years since the start of 2006 . Use the chain rule to estimate \(\left.\frac{d S}{d P}\right|_{t=2} .\) What does the answer tell you about home sales and prices? HINT [See Quick Example 2 on page 828.]
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