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Chapter 23: Problem 14

Find the derivative of each of the given functions. $$y=(1-6 x)^{4}$$

Short Answer

Expert verified
The derivative of the function is \( \frac{dy}{dx} = -24(1-6x)^{3} \).

Step by step solution

01

Recognize the Function Type

The given function is of the form \( y = (1-6x)^4 \), which is typically solved using the chain rule for derivatives. The function inside the parentheses is a linear expression \( u = 1-6x \), raised to the 4th power.
02

Apply the Chain Rule

The chain rule states that the derivative of \( y = (u)^n \) is \( n dot (u)^{n-1} dot \frac{du}{dx} \). Here, \( u = 1-6x \) and \( n = 4 \), so start by finding the derivative of \( u \) with respect to \( x \).
03

Differentiate the Inner Function

Take the derivative of \( u = 1-6x \). This gives \( \frac{du}{dx} = -6 \) because the derivative of a constant is 0 and the derivative of \(-6x\) is \(-6\).
04

Substitute in the Chain Rule Formula

Replace \( n \), \( u \), and \( \frac{du}{dx} \) in the chain rule formula: \( \frac{dy}{dx} = 4 dot (1-6x)^{4-1} dot (-6) \).
05

Simplify the Expression

Simplify the expression: \( \frac{dy}{dx} = 4 dot (1-6x)^{3} dot (-6) = -24(1-6x)^{3} \).

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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 tool in calculus used for finding the derivative of composite functions. A composite function is essentially a function within another function. In mathematical terms, if you have a function composed like this, \( y = f(g(x)) \), you use the chain rule to differentiate it.
  • The chain rule formula states that the derivative of \( y = f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \).
  • This means you first differentiate the outer function assuming the inner function is intact, and then multiply by the derivative of the inner function.
In our exercise, the function \( y = (1-6x)^4 \) is a perfect example. Here, the outer function is \( u^4 \) and the inner function is \( u = 1-6x \). By applying the chain rule, we first differentiate \( u^4 \) with respect to \( u \) and multiply it by the derivative of \( u = 1-6x \) with respect to \( x \).
Differentiation
Differentiation is a core process in calculus. It is used to find how a function changes at any given point. When you differentiate a function, you are essentially computing its derivative.
  • Derivatives tell you the rate at which a function is changing at any given point.
  • They are used extensively in various fields such as physics, engineering, and economics to model and predict changes.
In the given exercise, differentiating the function \( (1-6x)^4 \) involves two key differentiation steps. First, we differentiate the outer function \( u^4 \) while treating \( u \) as if it were a single variable. Then, we differentiate the inner function \( u=1-6x \), which results in \( -6 \) because the derivative of a constant is zero and \(-6x\) becomes \(-6\). The combination of these processes provides us with the overall derivative.
Linear Expression
A linear expression is a mathematical expression involving a linear function. Linear functions are those with a constant rate of change and can be written in the form \( y = mx + b \).
  • Linear expressions have no exponents higher than one, so they graph as straight lines.
  • The key feature is their coefficient, which tells you how steep the line is.
For our function, \( u = 1 - 6x \) is referred to as a linear expression. This means it will have a constant slope, which in this case is \(-6\). Since the derivative of a linear expression is simply its coefficient, differentiating \( u = 1 - 6x \) results in \(-6\). Linear expressions make applying the chain rule simpler because their derivatives are straightforward.
Calculus
Calculus is a branch of mathematics focusing on change and motion; it is divided into two main areas: differentiation and integration. Calculus helps us understand changes between values that are related by a function.
  • Differentiation lets us find the gradient of a curve, which is essentially its rate of change.
  • Integration, conversely, helps us find the total accumulation of quantities, such as areas under curves.
The problem we are tackling belongs to the differentiation part of calculus. Differentiation involves using rules, like the chain rule, to find the derivatives of functions. Understanding calculus and its principles not only allows us to solve real-world problems but also enhances analytical skills in observing how quantities evolve over time.

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Most popular questions from this chapter

Find the second derivative of each of the given functions. $$u=\frac{v^{2}}{4 v+15}$$

Solve the given problems. The cross section of a hill can be approximated by the curve of \(y=0.3 x-0.00003 x^{3}\) from \(x=0 \mathrm{m}\) to \(x=100 \mathrm{m} .\) The top of the hill is level. How high is the hill?

Find the derivative of each of the given functions. $$s=3\left(8 t^{2}-7\right)^{6}$$

Find \(d y / d x\) by differentiating implicitly. When applicable, express the result in terms of \(x\) and \(y\). $$6 x-3 y=4$$

Evaluate the indicated limits algebraically as in Examples \(10-14\). Change the form of the function where necessary. $$\lim _{h \rightarrow 0} \frac{(4+h)^{2}-16}{h}$$
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