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Chapter 23: Problem 6

In Exercises \(3-26,\) find the derivative of each of the functions by using the definition. $$y=2.3-5 x$$

Short Answer

Expert verified
The derivative of the function \( y = 2.3 - 5x \) is \( y' = -5 \).

Step by step solution

01

Recall the Definition of Derivative

The derivative of a function \( f(x) \) at a point \( x \) is defined as \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \). We will apply this definition to the function \( y = 2.3 - 5x \).
02

Express \( f(x+h) \)

Substitute \( x+h \) into the function \( y = 2.3 - 5x \) to get \( y = 2.3 - 5(x + h) \). Simplify this expression to obtain \( f(x+h) = 2.3 - 5x - 5h \).
03

Formulate the Difference Quotient

Substitute \( f(x+h) \) and \( f(x) \) into the difference quotient formula: \( \frac{f(x+h) - f(x)}{h} = \frac{(2.3 - 5x - 5h) - (2.3 - 5x)}{h} \).
04

Simplify the Difference Quotient

Simplify the expression \( \frac{(2.3 - 5x - 5h) - (2.3 - 5x)}{h} \). This simplifies to \( \frac{-5h}{h} \).
05

Apply the Limit

In the expression \( \frac{-5h}{h} \), cancel out \( h \) from the numerator and the denominator to obtain \( -5 \). Take the limit as \( h \to 0 \): \( \lim_{h \to 0} (-5) = -5 \). Thus, the derivative \( y' = -5 \).

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Key Concepts

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

Definition of Derivative
The concept of a derivative is essential in calculus, allowing us to understand how a function changes at any given point. Think of it like measuring the speed of a car at a specific moment. In mathematical terms, the derivative tells us the rate at which one quantity changes with respect to another. If you're trying to find the derivative of a function, you use the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Here's how to think about it:
  • Function Change: This measures how much the function's output changes as the input changes slightly (by \( h \)).
  • Limit Concept: By letting \( h \) approach zero, we're finding the very small changes, or the real-time rate of change.
This mathematical tool enables us to delve deeply into understanding behavior such as slopes of curves and motion dynamics.
Difference Quotient
At the heart of understanding derivatives is the difference quotient. This is essentially the formula used in the definition of a derivative. Imagine it as calculating an average rate of change, very similar to finding the slope between two points on a graph.
The formula goes like this:
  • \( \frac{f(x+h) - f(x)}{h} \)
This expression helps measure:
  • Output Change: \( f(x+h) \) minus \( f(x) \) shows how much the function values change.
  • Input Step: The change, \( h \), shows how far apart these points are.
The smaller the \( h \), the closer we get to the actual slope or rate of change at a specific point. This is why the limit (as covered in the next section) is so crucial for precision.
Limit Process
The limit process is vital in calculus because it allows us to analyze behavior as we approach a certain point. It transforms the difference quotient from an average view to a precise insight.
Here's what happens in the limit process:
  • Cancelling \( h \): In our simplified difference quotient, we often get forms like \( \frac{-5h}{h} \). Cancelling \( h \) gives us \(-5\).
  • Approaching Zero: The limit \( \lim_{h \to 0} \) makes \( h \) infinitesimally small or close to zero.
This process lets us find the exact slope of a tangent line to a curve at any point. So, in the example given, when you compute \( \lim_{h \to 0} (-5) \), you find that the derivative is \(-5\). This constant derivative means the function has a constant rate of change, which aligns with it being a linear function.

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Most popular questions from this chapter

Solve the given problems by finding the appropriate derivatives.The deflection \(y\) (in \(\mathrm{m}\) ) of a \(5.00-\mathrm{m}\) beam as a function of the distance \(x\) (in \(\mathrm{m}\) ) from one end is \(y=0.0001\left(x^{5}-25 x^{2}\right) .\) Find the value of \(d^{2} y / d x^{2}\) (the rate of change at which the slope of the beam changes) where \(x=3.00 \mathrm{m}\).

Solve the given problems by using implicit differentiation. A computer is programmed to draw the graph of the implicit function \(\left(x^{2}+y^{2}\right)^{3}=64 x^{2} y^{2}\) (see Fig. 23.45 and Example 7 on page 607 ). Find the slope of a line tangent to this curve at (2.00,0.56) and at (2.00,3.07)

Solve the given problems by finding the appropriate derivatives.What is the instantaneous rate of change of the first derivative of \(y\) with respect to \(x\) for \(y=(1-2 x)^{4}\) for \(x=1 ?\)

Solve the given problems by finding the appropriate derivatives. Find the derivative of \(y=\frac{x^{2}-1}{x-1}\) by (a) the quotient rule, and (b) by first simplifying the function.

\(\lim _{x \rightarrow a^{-}} f(x)\) means to find the limit as x approaches a from the left only, and \(\lim _{x \rightarrow a^{+}} f(x)\) means to find the limit as \(x\) approaches a from the right only. These are called one-sided limits. Solve the following problems. $$\text { Find } \lim _{x \rightarrow 4^{-}} x \sqrt{16-x^{2}}$$
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