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Hyperbolic functions are analogs to trigonometric functions but for a hyperbola, just as sine and cosine are for a circle. One of the basic hyperbolic functions is the hyperbolic tangent, represented as \( \tanh x \). The function \( \tanh x \) is defined using exponential functions: \( \tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}} \). This function has some unique properties, much like its trigonometric counterpart, the tangent function. Some notable characteristics include:

• Range: The values of \( \tanh x \) range from -1 to 1.
• Symmetry: \( \tanh x \) is an odd function, meaning \( \tanh(-x) = -\tanh(x) \).
• Zeros: It crosses the x-axis at \( x = 0 \).

Understanding these properties is essential, especially when dealing with the calculation of derivatives and solving problems involving hyperbolic tangent functions.

The derivative is a fundamental concept in calculus representing the rate of change of a function. For the function \( f(x) = \tanh x \), its derivative is \( \text{sech}^2 x \). The hyperbolic secant function, denoted as \( \text{sech} x \), is defined as \( \text{sech} x = \frac{1}{\cosh x} \). This derivative tells us how \( \tanh x \) changes at different points along the x-axis.
Calculating the derivative is the first crucial step in finding the equation of a tangent line, as it provides the slope. The expression \( \text{sech}^2 x \) indicates that the rate of change for \( \tanh x \) is linked directly to the hyperbolic cosine function, \( \cosh x \), emphasizing their interconnectedness in hyperbolic functions.

Finding the slope of the tangent line involves evaluating the derivative at a specific x-value. In this exercise, you substitute \( x = -\ln 3 \) into the derivative \( f'(x) = \text{sech}^2 x \). This step determines how steep or flat the tangent line will be at that point.

Here's how it was calculated:

• Recognize that \( \cosh(-x) = \cosh(x) \).
• Calculate \( \cosh(\ln 3) = \frac{e^{\ln 3} + e^{-\ln 3}}{2} = \frac{3 + \frac{1}{3}}{2} = \frac{5}{3} \).
• Then, find \( \text{sech}(\ln 3) = \frac{3}{5} \).
• Finally, \( \text{sech}^2(\ln 3) = \left(\frac{3}{5}\right)^2 = \frac{9}{25} \).

This slope \( \frac{9}{25} \) will be used in the equation of the tangent line.

The point-slope form is a method of writing equations of lines using a known point on the line and the slope. It is a convenient way to find the equation of a tangent line once you've determined the slope. The general form is:\[ y - y_1 = m(x - x_1) \]where:

• \( (x_1, y_1) \) is a point on the line, which in this case is \((-\ln 3, -\frac{4}{5})\).
• \( m \) is the slope, which we've found to be \( \frac{9}{25} \).

Substitute these values into the formula to get:\[ y + \frac{4}{5} = \frac{9}{25}(x + \ln 3) \]After simplification, the equation of the tangent line becomes:\[ y = \frac{9}{25}(x + \ln 3) - \frac{4}{5} \] This equation represents the precise line that just touches \( f(x) = \tanh x \) at the given point.