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Chapter 2: Problem 3

Find the derivative of the function. $$ y=8 $$

Short Answer

Expert verified
The derivative of the function \(y=8\) is \(0\).

Step by step solution

01

Identify the function

The given function is \(y=8\). This is a constant function, meaning it does not depend on any variables.
02

Apply the rule for differentiation of a constant

The rule for finding the derivative of a constant is that the derivative is always 0. This is because a derivative measures how a function changes as its inputs change, and a constant function does not change.
03

Result of differentiation

Therefore, the derivative of the function \(y=8\) is \(dy/dx = 0\).

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Key Concepts

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

Differentiation rules
Differentiation is the process in calculus that allows us to determine the rate at which a function changes. Essentially, it's like measuring the steepness of a curve at any given point. Differentiation rules are guidelines that simplify this process. One of the simplest rules is the **derivative of a constant function**.
  • For any constant value, the rate of change is zero. Therefore, its derivative is zero.
  • This is often one of the first rules students learn because it's easy to remember and intuitive.
Another important rule to understand is the **power rule**, which applies to functions of the form \(x^n\). However, since a constant can be thought of as \(x^0\), even here, applying the power rule confirms the derivative is zero. Differentiation rules like these are foundational in calculus, providing tools to tackle more complex functions later on.
Constant function
A constant function is one of the simplest types of functions we encounter in mathematics. It is defined as \[ y = c \]where \(c\) is a constant number. What makes constant functions unique is that they do not vary with changes in the independent variable. This means that no matter what value you substitute into the function, the output remains the same - in this case, always 8.
  • Constant functions have a derivative of zero, as they do not change regardless of any inputs.
  • Graphically, a constant function is represented as a horizontal line.
Understanding constant functions is crucial as it establishes a baseline for more complex functions.
Calculus basics
Calculus is a branch of mathematics that focuses on change and motion. It is divided mainly into two areas: differential calculus and integral calculus. At its core, calculus helps us understand how things change over time and space.
  • **Differential calculus** involves finding the rate of change. This is achieved through differentiation, where we learn about derivatives.
  • **Integral calculus** is about accumulation of quantities, which is essentially the reverse of differentiation.
Together, these two branches allow us to explore and model the world around us mathematically. Basic concepts like the derivative of a constant are stepping stones to more advanced topics. Mastering these basics sets the groundwork for tackling challenging problems in physics, engineering, economics, and beyond.

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Most popular questions from this chapter

Verify each differentiation formula. (a) \(\frac{d}{d x}[\arctan u]=\frac{u^{\prime}}{1+u^{2}}\) (b) \(\frac{d}{d x}[\operatorname{arccot} u]=\frac{-u^{\prime}}{1+u^{2}}\) (c) \(\frac{d}{d x}[\operatorname{arcsec} u]=\frac{u^{\prime}}{|u| \sqrt{u^{2}-1}}\) (d) \(\frac{d}{d x}[\arccos u]=\frac{-u^{\prime}}{\sqrt{1-u^{2}}}\) (e) \(\frac{d}{d x}[\operatorname{arccsc} u]=\frac{-u^{\prime}}{|u| \sqrt{u^{2}-1}}\)

The area of a square with sides of length \(s\) is given by \(A=s^{2} .\) Find the rate of change of the area with respect to \(s\) when \(s=4\) meters.

Find the derivative of the function. \(f(t)=\frac{3^{2 t}}{t}\)

(a) Find an equation of the normal line to the ellipse \(\frac{x^{2}}{32}+\frac{y^{2}}{8}=1\) at the point (4,2) . (b) Use a graphing utility to graph the ellipse and the normal line. (c) At what other point does the normal line intersect the ellipse?

Let \(u\) be a differentiable function of \(x\). Use the fact that \(|u|=\sqrt{u^{2}}\) to prove that \(\frac{d}{d x}[|u|]=u^{\prime} \frac{u}{|u|}, \quad u \neq 0\).
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