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Chapter 23: Problem 20

Find the derivative of each of the functions by using the definition. $$y=\frac{5 x}{x-1}$$

Short Answer

Expert verified
The derivative is \( y' = \frac{-5}{(x-1)^2} \).

Step by step solution

01

Understand the Definition of Derivative

The derivative of a function \( f(x) \) at a point \( x = c \) is defined as \( f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} \). In this problem, we must find the derivative of \( y = \frac{5x}{x-1} \) using this definition.
02

Substitute and Simplify

First, substitute \( f(x) = \frac{5x}{x-1} \) into the limit definition: \[ f'(x) = \lim_{h \to 0} \frac{\frac{5(x+h)}{(x+h)-1} - \frac{5x}{x-1}}{h}. \] Now simplify the expression.
03

Find a Common Denominator

To simplify, find a common denominator for the fraction inside the limit: \((x-1)((x+h)-1) = (x-1)(x+h-1) \). Then the expression becomes: \[ \frac{5(x+h)(x-1) - 5x(x+h-1)}{h(x-1)(x+h-1)}. \] Simplify the numerator.
04

Simplify the Numerator

Expand and simplify the numerator: \(5x^2 - 5x + 5hx - 5h - (5x^2 + 5hx - 5x) = -5h \). Thus the expression becomes: \[ \frac{-5h}{h(x-1)(x+h-1)}. \]
05

Cancel out the \( h \) terms

Cancel \( h \) from the numerator and the denominator: \[ \frac{-5}{(x-1)(x+h-1)}. \] Now, find the limit as \( h \to 0 \).
06

Evaluate the Limit

Evaluate the limit: \[ \lim_{h \to 0} \frac{-5}{(x-1)(x+h-1)} = \frac{-5}{(x-1)x}. \] We have the derivative of the function.

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Key Concepts

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

Derivative
In calculus, a derivative is a fundamental concept that measures how a function changes as its input changes. Understanding derivatives involves grasping several key ideas:
  • The derivative of a function at a specific point provides the rate of change or the slope of the tangent line at that point.
  • Derivatives can be thought of as the "instantaneous" rate of change, a concept that becomes clearer when we consider how small changes in the input affect the function's output.
In simpler terms, if you imagine a curve on a graph, the derivative at any given point is like knowing how steep that curve is precisely at that instant. This is immensely useful in various fields such as physics for determining velocity or in economics for calculating marginal costs. Learning to find derivatives gives us powerful tools to analyze and interpret real-world phenomena.
Limit Definition
The limit definition of a derivative is the mathematical way to precisely define what a derivative is. It is represented as:\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. \]This equation may look complex, but it breaks down into manageable parts:
  • The expression \( f(x+h) - f(x) \) measures how much the function \( f(x) \) changes as you move a small amount \( h \) from \( x \).
  • Dividing that change by \( h \) gives you an average rate of change over that tiny interval.
  • Taking the limit as \( h \) approaches zero fine-tunes this rate to the exact point where you want the derivative. Essentially, it zeros in on how the function changes instantaneously.
The elegance of this definition lies in its ability to apply universally to any function type, including complex ones.
Simplification
Simplification is a crucial step in solving calculus problems, including finding derivatives. In this context, it means reducing complex mathematical expressions to their simplest form. Here is how simplification helps in derivative calculation:
  • Simplifying expressions often involves factorizing, combining like terms, and using algebraic identities to make calculations more manageable.
  • Effective simplification ensures that unnecessary complexities don't interfere with the core calculation, making it both more accurate and easier to understand.
  • In the example provided, simplifying helps us to remove excess variables and terms so that we can clearly see where to cancel terms, like the \( h \) in the numerator and denominator.
Using simplification is like clearing a path. It helps make sure the steps are as clear as possible and the solution is correctly obtained.
Rational Functions
Rational functions are fractions involving polynomials in the numerator and denominator. They come with some specific characteristics that influence derivative calculations:
  • When differentiating rational functions, it's common to encounter complex fractions, as seen in the exercise with \( \frac{5x}{x-1} \). These require careful handling to ensure accuracy.
  • Rational functions often involve vertical asymptotes, where the function increases or decreases without bound, which requires precise calculus techniques.
  • The derivative of a rational function helps show trends and changes in progression over the domain of interest, useful for functions of this nature where the behavior can shift dramatically around asymptotes or intercepts.
Understanding rational functions in the context of derivatives helps you realize the delicate interplay between algebraic expressions and calculus concepts.

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Most popular questions from this chapter

Evaluate the indicated limits by direct evaluation as in Examples \(10-14 .\) Change the form of the function where necessary. $$\lim _{x \rightarrow 0} \frac{6 x^{2}+x}{x}$$

Solve the given problems by using implicit differentiation.Show that two tangents to the curve \(x^{2}+x y+y^{2}=7\) at the points where it crosses the \(x\) -axis are parallel.

\(\lim _{x \rightarrow a^{-}} f(x)\) 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. Is there a difference between \(\lim _{x \rightarrow 2^{-}} \frac{1}{x-2}\) and \(\lim _{x \rightarrow 2^{+}} \frac{1}{x-2} ?\)

Solve the given problems involving limits. For a quadratic equation \(a+b x+c=0,\) the solutions are \(x_{1}=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\) and \(x_{2}=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a} .\) Show that, as \(a \rightarrow 0, x_{1} \rightarrow-b / c\) and \(x_{2} \rightarrow \infty .\) (Hint: rationalize numerators.) (Note that \(-b / c\) is the root of the equation \(b x+c=0 .)\)

Solve the given problems involving limits. A \(5-\Omega\) resistor and a variable resistor of resistance \(R\) are placed in parallel. The expression for the resulting resistance \(R_{T}\) is given by \(R_{T}=\frac{5 R}{5+R} .\) Determine the limiting value of \(R_{T}\) as \(R \rightarrow \infty\)
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