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Magazine Chapter 6: Problem 26
Find the equation of the line tangent to the function at the given \(x\) -value. \(f(x)=\operatorname{sech}^{2} x\) at \(x=\ln 3\)
Short Answer
Expert verified
The equation of the tangent line is \( y - \frac{9}{25} = -\frac{72}{125}(x - \ln 3) \).
Step by step solution
01
Find the Derivative
The tangent line at a point requires the derivative of the function. Given the function \( f(x) = \operatorname{sech}^2 x \), where \( \operatorname{sech} x = \frac{1}{\cosh x} \), we start by finding the derivative using the chain rule: \( f'(x) = -2 \cdot \operatorname{sech} x \cdot \operatorname{sech} x \cdot \tanh x = -2 \cdot \operatorname{sech}^2 x \cdot \tanh x \).
02
Evaluate the Derivative at the Point
Substitute \( x = \ln 3 \) into the derivative to find the slope of the tangent line at this point. First, compute \( \cosh (\ln 3) \) and \( \sinh (\ln 3) \) using: \( \cosh(\ln 3) = \frac{3 + \frac{1}{3}}{2} = \frac{10}{6} = \frac{5}{3} \) and \( \sinh(\ln 3) = \frac{3 - \frac{1}{3}}{2} = \frac{8}{6} = \frac{4}{3} \). Hence, \( \operatorname{sech}(\ln 3) = \frac{1}{\frac{5}{3}} = \frac{3}{5} \) and \( \tanh(\ln 3) = \frac{4}{5} \). Substitute these into the derivative: \( f'(\ln 3) = -2 \left(\frac{3}{5}\right)^2 \cdot \frac{4}{5} = -2 \cdot \frac{9}{25} \cdot \frac{4}{5} = -\frac{72}{125} \).
03
Find the Function Value at the Given Point
Find \( f(\ln 3) = \operatorname{sech}^2(\ln 3) = \left(\frac{3}{5}\right)^2 = \frac{9}{25} \).
04
Use Point-Slope Formula to Write Tangent Line Equation
Using the slope from Step 2 and the function value from Step 3, apply the point-slope form of a line equation: \( y - f(\ln 3) = f'(\ln 3)(x - \ln 3) \). Substitute the values: \( y - \frac{9}{25} = -\frac{72}{125}(x - \ln 3) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
To find the tangent line's equation to a function involving hyperbolic functions, such as \( f(x) = \operatorname{sech}^2 x \), the chain rule is an essential tool. The chain rule helps us differentiate composite functions, and it follows this principle:
- If you have a function \( g(f(x)) \), its derivative is \( g'(f(x)) \cdot f'(x) \).
- Differentiating \( [\operatorname{sech} x]^2 \) involves the outer function \( u^2 \), where \( u = \operatorname{sech} x \).
- The derivative of \( u^2 \) is \( 2u \), so we multiply by the derivative of the inside function \( \operatorname{sech} x \).
- The derivative of \( \operatorname{sech} x \) is \( -\operatorname{sech}(x) \cdot \tanh(x) \).
- Putting it all together, the chain rule gives us: \( f'(x) = -2 \cdot \operatorname{sech} x \cdot \operatorname{sech} x \cdot \tanh x \).
Hyperbolic Functions
Hyperbolic functions are analogs of the trigonometric functions but for a hyperbola. They provide valuable tools in calculus and appear frequently in real-world scenarios. The key hyperbolic functions involved here are:
- \( \operatorname{cosh}(x) \): Represents the hyperbolic cosine. Defined as \( \frac{e^x + e^{-x}}{2} \).
- \( \operatorname{sinh}(x) \): Represents the hyperbolic sine. Defined as \( \frac{e^x - e^{-x}}{2} \).
- \( \operatorname{sech}(x) \): This is the hyperbolic secant, \( \operatorname{sech}(x) = \frac{1}{\cosh(x)} \).
- \( \operatorname{tanh}(x) \): The hyperbolic tangent, \( \operatorname{tanh}(x) = \frac{\sinh(x)}{\cosh(x)} \).
- The derivative of \( \operatorname{sech}(x) \) is \( -\operatorname{sech}(x) \cdot \operatorname{tanh}(x) \).
- The derivative of \( \operatorname{tanh}(x) \) is \( \operatorname{sech}^2(x) \).
Derivative Evaluation
Evaluating the derivative at a specific point is a major step in finding the tangent line equation. After finding the derivative, which was \( -2 \cdot \operatorname{sech}^2 x \cdot \tanh x \) for our given function, we need to substitute \( x = \ln 3 \). Steps to evaluate the derivative at \( x = \ln 3 \):
- First, compute \( \cosh(\ln 3) = \frac{3 + \frac{1}{3}}{2} = \frac{5}{3} \).
- Next, calculate \( \sinh(\ln 3) = \frac{3 - \frac{1}{3}}{2} = \frac{4}{3} \).
- Thus, \( \operatorname{sech}(\ln 3) = \frac{1}{\frac{5}{3}} = \frac{3}{5} \).
- For \( \operatorname{tanh}(\ln 3) = \frac{4}{5} \).
- \( f'(\ln 3) = -2 \left(\frac{3}{5}\right)^2 \cdot \frac{4}{5} = -\frac{72}{125} \).
Point-Slope Form
The point-slope form is a method to write the equation of a tangent line at a particular point. If you know the point \((x_1, y_1)\) on the function and the slope \(m\), the equation is:\[ y - y_1 = m(x - x_1) \]In our problem:
- First, we found the function value at the point: \( f(\ln 3) = \operatorname{sech}^2(\ln 3) = \frac{9}{25} \).
- The slope from our derivative evaluation was \( m = -\frac{72}{125} \).
- Plug in \( y_1 = \frac{9}{25} \), \( m = -\frac{72}{125} \), and \( x_1 = \ln 3 \):
- \( y - \frac{9}{25} = -\frac{72}{125}(x - \ln 3) \).
