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Chapter 3: Problem 30

Find the equation of the tangent line to the curve \(y=(x+1) /(x-1)\) at the point \((2,3)\).

Short Answer

Expert verified
The equation of the tangent line is \(y = -2x + 7\)

Step by step solution

01

Find the derivative of the function

Given the function \(y = \frac{x+1}{x-1}\), find the derivative \(y'\). Use the quotient rule which states \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Let \(u = x+1\) and \(v = x-1\). Calculate \(u'\) and \(v'\) and then substitute into the quotient rule formula.
02

Apply the quotient rule

Using \(u' = 1\) and \(v' = 1\), the derivative \(y'\) is:\[ y' = \frac{(x-1)(1) - (x+1)(1)}{(x-1)^2} = \frac{x-1-x-1}{(x-1)^2} = \frac{-2}{(x-1)^2} \]
03

Evaluate the derivative at the given point

Substitute \(x = 2\) into the derivative \(y' = \frac{-2}{(x-1)^2}\).\[ y'|_{x=2} = \frac{-2}{(2-1)^2} = \frac{-2}{1} = -2 \]
04

Use the point-slope form to find the equation of the tangent line

The point-slope form of the equation of a line is \(y - y_1 = m(x - x_1)\). Here, \(m = -2\), \(x_1 = 2\), and \(y_1 = 3\). Substitute these values into the formula:\[ y - 3 = -2(x - 2) \]
05

Simplify the equation

Simplify the equation to get the final answer:\[ y - 3 = -2x + 4 \]\[ y = -2x + 7 \]

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Key Concepts

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

Quotient Rule
The quotient rule is a method used in calculus to find the derivative of a function that is the quotient of two other functions. If you have a function in the form \( \frac{u}{v} \), the quotient rule states that its derivative is given by \( \frac{u'v - uv'}{v^2} \). Here, \( u \) and \( v \) are both functions of \( x \), with \( u' \) and \( v' \) representing their derivatives.
Derivative
A derivative represents the rate at which a function is changing at any given point. In simpler terms, it gives you the slope of the tangent line to the function at a specific point. For our problem, we start with the function \( y = \frac{x+1}{x-1} \) and first identify \( u \) and \( v \):
  • \( u = x + 1 \)
  • \( v = x - 1 \)
The derivatives are:
  • \( u' = 1 \)
  • \( v' = 1 \)
Plug these into the quotient rule:
\[ y' = \frac{(x-1)(1) - (x+1)(1)}{(x-1)^2} = \frac{x-1-x-1}{(x-1)^2} = \frac{-2}{(x-1)^2} \] Understanding derivatives is crucial to find out how the function behaves at a specific point.
Point-Slope Form
To find the equation of the tangent line, we use the point-slope form of a line, which is:
\[ y - y_1 = m(x - x_1) \] Here, \( m \) is the slope, and \( (x_1, y_1) \) is a point on the line. Based on our earlier steps, we found the slope to be \( -2 \) at the point \( (2, 3) \). Substituting these values into the point-slope form gives:
\[ y - 3 = -2(x - 2) \] Simplify to get the equation of the tangent line:
\[ y = -2x + 7 \] This is the final tangent line equation for the curve \( y = \frac{x+1}{x-1} \) at the point \( (2, 3) \).

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Most popular questions from this chapter

Demand Equation Suppose that \(x\) and \(y\) represent the amounts of two basic inputs for a production process and $$ 10 x^{1 / 2} y^{1 / 2}=600 . $$ Find \(\frac{d y}{d x}\) when \(x=50\) and \(y=72\).

If \(f(x)\) is a function whose derivative is \(f^{\prime}(x)=\frac{1}{1+x^{2}}\), find the derivative of \(\frac{f(x)}{1+x^{2}}\).

Suppose that \(x\) and \(y\) are both differentiable functions of \(t\) and are related by the given equation. Use implicit differentiation with respect to \(t\) to determine \(\frac{d y}{d t}\) in terms of \(x, y\), and \(\frac{d x}{d t}\). $$x^{4}+y^{4}=1$$

Suppose that \(x\) and \(y\) are related by the given equation and use implicit differentiation to determine \(\frac{d y}{d x}\). $$x^{2}-y^{2}=1$$

A point is moving along the graph of \(x^{3} y^{2}=200 .\) When the point is at \((2,5)\), its \(x\) -coordinate is changing at the rate of \(-4\) units per minute. How fast is the \(y\) -coordinate changing at that moment?
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