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Magazine Chapter 4: Problem 87
$$ \text { Differentiate. } $$ $$ y=\ln \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} $$
Short Answer
Expert verified
The derivative is \( y' = \frac{4}{\text{sinh}(2x)} \).
Step by step solution
01
Simplify the Argument
First, let's simplify the argument of the natural logarithm. Define \( u = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}} \).
02
Differentiate Using Chain Rule
To find the derivative \( y' \), utilize the chain rule. This requires differentiating the natural logarithm function and then the inner function \( u \). The chain rule states \( \frac{d}{dx} \big[ \text{ln}(u) \big] = \frac{1}{u} \frac{du}{dx} \).
03
Differentiate the Quotient
Now, apply the quotient rule to \( \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}} \). Let \( v = e^{x} - e^{-x} \) and \( w = e^{x} + e^{-x} \). The quotient rule states: \[ \frac{d}{dx} \bigg( \frac{v}{w} \bigg) = \frac{v'w - vw'}{w^2} \]
04
Compute the Derivatives
Calculate \( v' \) and \( w' \). Since \( v = e^{x} - e^{-x} \), we have \( v' = e^{x} + e^{-x} \). Since \( w = e^{x} + e^{-x} \), we have \( w' = e^{x} - e^{-x} \).
05
Substitute Derivatives into Quotient Rule
Substitute \( v, v', w, \) and \( w' \) into the quotient rule: \[ \frac{du}{dx} = \frac{(e^{x} + e^{-x})(e^{x} + e^{-x}) - (e^{x} - e^{-x})(e^{x} - e^{-x})}{(e^{x} + e^{-x})^2} \]
06
Simplify the Derivative
Simplify the expression: \[ \frac{du}{dx} = \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{(e^{x} + e^{-x})^2} = \frac{4}{(e^{x} + e^{-x})^2} \]
07
Combine Results
Combine the results from previous steps: \[ y' = \frac{1}{\frac{e^{x} - e^{-x}}{e^{x} + e^{-x}}} \times \frac{4}{(e^{x} + e^{-x})^2} = \frac{4(e^{x} + e^{-x})}{(e^{x} - e^{-x})(e^{x} + e^{-x})^2} = \frac{4}{(e^{x} - e^{-x})(e^{x} + e^{-x})} \]
08
Final Simplification
Simplify further to the final form: \[ y' = \frac{4}{e^{2x} - e^{-2x}} = \frac{4}{\text{sinh}(2x)} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Natural Logarithm Derivative
To differentiate the natural logarithm function, we use the fact that the derivative of \(\text{ln}(u)\) with respect to \(x\) is given by \(\frac{1}{u} \frac{du}{dx}\). This tells us that we'll need to break our job into two parts: first, finding the derivative of the inner function \(u\), and second, using that result to find the derivative of the overall function. Differentiating natural logarithms often involves simplifying the expression inside the log (the argument), which in many cases can be a quotient.
Chain Rule
The chain rule is crucial in calculus for differentiating compositions of functions. Mathematically, the chain rule is represented as \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \times g'(x)\). In graphical terms, the chain rule can be thought of as analyzing the variations in an outer function in response to changes in an inner function. When differentiating \(\text{ln}(u)\) with \(\text{u = the argument inside}\), we use the chain rule to get \(\frac{1}{u} \frac{du}{dx}\), where \(u = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}})\). This simplification allows us to keep track of different variables and their respective derivatives neatly.
Quotient Rule
Differentiating a quotient \( \frac{v}{w} \) involves the quotient rule, which states: \[ \frac{d}{dx} \bigg( \frac{v}{w} \bigg) = \frac{v'w - vw'}{w^2} \]. Here, \(v\) and \(w\) are functions of \(x\). For instance, if \(v = e^{x} - e^{-x}\) and \(w = e^{x} + e^{-x}\), then their derivatives, \(v' = e^{x} + e^{-x}\) and \(w' = e^{x} - e^{-x}\) respectively, are computed and substituted back into the formula to get the derivative of the overall function. This rule is helpful for expressions where division is involved.
Hyperbolic Functions
Hyperbolic functions, such as \( \text{sinh}(x) \) and \( \text{cosh}(x) \), often appear in calculus problems. These functions are analogs of the trigonometric functions but for a hyperbola, just as trigonometric functions are for a circle. For example, \( \text{sinh}(x) = \frac{e^x - e^{-x}}{2} \) and \( \text{cosh}(x) = \frac{e^x + e^{-x}}{2} \). In our solution, recognizing that \( e^{2x} - e^{-2x} \) is equivalent to \( \text{sinh}(2x) \) simplifies the final derivative to \( y' = \frac{4}{\text{sinh}(2x)} \). Noticing these hyperbolic relationships can make complicated algebra much more manageable.
