Problem 62
Question
Find an equation of the tangent line to the graph of the function at the given point. \(f(x)=\left(\frac{x+1}{x-1}\right)^{2} ;(3,4)\)
Step-by-Step Solution
Verified Answer
The equation of the tangent line to the graph of the function \(f(x) = (\frac{x+1}{x-1})^{2}\) at the point \((3,4)\) is \(y = -4x + 16\).
1Step 1: Find the derivative of the function
To find the derivative of the function \(f(x)=(\frac{x+1}{x-1})^2\), we will use the chain rule, which states that if we have a function in the form \(f(g(x))\), then the derivative is given by \[f'(g(x)) \cdot g'(x).\] In this case, let \(g(x) = \frac{x+1}{x-1}\) and \(f(u)=u^2\), where \(u=g(x)\). Now, we will find the derivatives of \(f(u)\) and \(g(x)\):
1. For \(f(u)=u^2\), the derivative with respect to \(u\) is \(f'(u) = 2u\).
2. To find the derivative of \(g(x) =\frac{x+1}{x-1}\), we will use the quotient rule, which states that \[\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2},\] where \(u(x) = x + 1\) and \(v(x) = x - 1\).
Now, find their derivatives:
1. The derivative of \(u(x)=x+1\) is \(u'(x)=1\).
2. The derivative of \(v(x)=x-1\) is \(v'(x)=1\).
Apply the quotient rule for \(g'(x)\):
\[g'(x)=\frac{u'v - uv'}{v^2}=\frac{(1)(x-1)-(x+1)(1)}{(x-1)^2}=\frac{x-1-x-1}{(x-1)^2}=\frac{-2}{(x-1)^2}.\]
Now, using the chain rule, we have:
\[f'(x) = f'(g(x)) \cdot g'(x) = 2 \cdot \frac{x+1}{x-1} \cdot \left(-\frac{2}{(x-1)^2}\right).\]
2Step 2: Evaluate the derivative at the given point
We are given the point \((3,4)\), so we need to evaluate \(f'(x)\) at \(x=3\) to find the slope of the tangent line at this point:
\[f'(3) = 2 \cdot \frac{3+1}{3-1} \cdot \left(-\frac{2}{(3-1)^2}\right) = 2 \cdot \frac{4}{2} \cdot \left(-\frac{2}{4}\right) = 4 \cdot (-1) = -4.\]
So, the slope of the tangent line at the given point is -4.
3Step 3: Use the point-slope form to find the equation of the tangent line
Now that we have the slope of the tangent line (-4) and the point \((3, 4)\), we can use the point-slope form to find the equation of the tangent line. The point-slope form is given by:
\[y - y_1 = m(x - x_1),\]
where \(m\) is the slope, and \((x_1, y_1)\) is the given point.
Plug in the values for \(m\), \(x_1\), and \(y_1\):
\[y - 4 = -4(x - 3).\]
Now, simplify and solve for \(y\):
\[y - 4 = -4x + 12 \Rightarrow y = -4x + 12 + 4.\]
4Step 4: Write down the equation of the tangent line
The equation of the tangent line to the graph of the function at the given point is:
\[y = -4x + 16.\]
Key Concepts
DerivativeChain RuleQuotient Rule
Derivative
The concept of a derivative is foundational in understanding how functions change. In simple terms, the derivative of a function at a specific point provides the slope of the tangent line to the curve at that point. This means it tells us how steep the graph is and the direction it is heading.
When dealing with equations, the process of differentiation is how we find this slope. Differentiation involves finding the derivative by using rules and formulas that cater to the function's structure. For example, the power rule, product rule, and chain rule are tools we use to calculate derivatives. In the case of the function \( f(x) = \left( \frac{x+1}{x-1} \right)^2 \), the derivative gives us the slope of the tangent line.
When dealing with equations, the process of differentiation is how we find this slope. Differentiation involves finding the derivative by using rules and formulas that cater to the function's structure. For example, the power rule, product rule, and chain rule are tools we use to calculate derivatives. In the case of the function \( f(x) = \left( \frac{x+1}{x-1} \right)^2 \), the derivative gives us the slope of the tangent line.
Chain Rule
The chain rule is a powerful tool in calculus, especially useful when dealing with composite functions. A composite function is one where a function is applied within another function, like \(f(g(x))\). The chain rule helps us differentiate such compositions efficiently.
The rule states: If \( y = f(g(x)) \), then the derivative \( y' \) is \( f'(g(x)) \cdot g'(x) \). This means we first differentiate the outer function with respect to the inner function, then multiply by the derivative of the inner function. It's like peeling an onion; you work from the outermost layer to the innermost.
In our function \( f(x) = \left( \frac{x+1}{x-1} \right)^2 \), the outer function is \( u^2 \) and the inner function \( g(x) = \frac{x+1}{x-1} \). We find \( f'(u) = 2u \) and \( g'(x) = -\frac{2}{(x-1)^2} \), then multiply them together using the chain rule.
The rule states: If \( y = f(g(x)) \), then the derivative \( y' \) is \( f'(g(x)) \cdot g'(x) \). This means we first differentiate the outer function with respect to the inner function, then multiply by the derivative of the inner function. It's like peeling an onion; you work from the outermost layer to the innermost.
In our function \( f(x) = \left( \frac{x+1}{x-1} \right)^2 \), the outer function is \( u^2 \) and the inner function \( g(x) = \frac{x+1}{x-1} \). We find \( f'(u) = 2u \) and \( g'(x) = -\frac{2}{(x-1)^2} \), then multiply them together using the chain rule.
Quotient Rule
The quotient rule is essential for differentiating functions that are divided by each other, such as rational functions. If you have a function in the form \( \frac{u(x)}{v(x)} \), the quotient rule provides a formula to find its derivative.
The rule is given by:
This means you first find the derivative of the numerator \( u(x) \) and the derivative of the denominator \( v(x) \). Then, create a new fraction: the numerator is the difference between \( u'v \) and \( uv' \), and the new denominator is \( v^2 \). This carefully balances changes in the numerator and denominator to get the correct slope of the quotient.
For our function component \( g(x) = \frac{x+1}{x-1} \), the quotient rule helps find \( g'(x) = -\frac{2}{(x-1)^2} \). It carefully ensures we account for changes in both the numerator and denominator.
The rule is given by:
- \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \)
This means you first find the derivative of the numerator \( u(x) \) and the derivative of the denominator \( v(x) \). Then, create a new fraction: the numerator is the difference between \( u'v \) and \( uv' \), and the new denominator is \( v^2 \). This carefully balances changes in the numerator and denominator to get the correct slope of the quotient.
For our function component \( g(x) = \frac{x+1}{x-1} \), the quotient rule helps find \( g'(x) = -\frac{2}{(x-1)^2} \). It carefully ensures we account for changes in both the numerator and denominator.
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