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Magazine Chapter 2: Problem 431
Prove the formula for the derivative of \(y=\operatorname{sech}^{-1}(x)\) by differentiating \(x=\operatorname{sech}(y) .\) (Hint: Use hyperbolic trigonometric identities.)
Short Answer
Expert verified
The derivative of \( y = \operatorname{sech}^{-1}(x) \) is \( \frac{dy}{dx} = -\frac{1}{x \cdot \sqrt{1 - x^2}} \).
Step by step solution
01
Express the Given Function in Terms of y
Start with the given function: \( x = \operatorname{sech}(y) \). Recall that the hyperbolic secant function is defined as \( \operatorname{sech}(y) = \frac{2}{e^y + e^{-y}} \). Thus, we have: \( x = \frac{2}{e^y + e^{-y}} \).
02
Solve for y in Terms of x
Solve the equation \( x = \frac{2}{e^y + e^{-y}} \) for \( y \). Multiply both sides by \( e^y + e^{-y} \) to get: \( x(e^y + e^{-y}) = 2 \). This simplifies to: \( xe^y + xe^{-y} = 2 \).
03
Isolate \( e^y \)
Multiply the whole equation by \( e^y \) to eliminate \( e^{-y} \): \( xe^{2y} + x = 2e^y \). Rearrange to form a quadratic in terms of \( e^y \): \( xe^{2y} - 2e^y + x = 0 \).
04
Solve the Quadratic Equation for \( e^y \)
Use the quadratic formula \( e^y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = x \), \( b = -2 \), and \( c = x \) to find \( e^y \). This gives us \( e^y = \frac{2 \pm \sqrt{4 - 4x^2}}{2x} \). Simplify to get \( e^y = \frac{1 \pm \sqrt{1 - x^2}}{x} \). Since \( e^y \) must be positive, choose the positive solution: \( e^y = \frac{1 + \sqrt{1 - x^2}}{x} \).
05
Differentiate Implicitly to Find \( \frac{dy}{dx} \)
Differentiate the original equation \( x = \operatorname{sech}(y) \) implicitly: \( 1 = -\operatorname{sech}(y)\tanh(y) \frac{dy}{dx} \). Isolate \( \frac{dy}{dx} \) to get \( \frac{dy}{dx} = -\frac{1}{\operatorname{sech}(y)\tanh(y)} \).
06
Substitute Hyperbolic Trigonometric Identities
Recall that \( \operatorname{sech}^2(y) + \tanh^2(y) = 1 \) and hence \( \tanh(y) = \sqrt{1 - \operatorname{sech}^2(y)} = \sqrt{1 - x^2} \). Thus, \( \frac{dy}{dx} = -\frac{1}{x \cdot \sqrt{1 - x^2}} \), as the identities yield \( \operatorname{sech}(y) = x \) and \( \tanh(y) = \sqrt{1 - x^2} \).
07
Simplify the Expression
The derivative of \( y = \operatorname{sech}^{-1}(x) \) is \( \frac{dy}{dx} = -\frac{1}{x \cdot \sqrt{1 - x^2}} \), which corresponds to the expression derived from step 6. This completes the proof.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Implicit Differentiation
Implicit differentiation is a particularly useful technique for differentiating equations where the variable you want to differentiate with respect to, say \( x \), is mixed with another variable, say \( y \), that also depends on \( x \).
In this problem, we are tasked with finding the derivative of \( y = \operatorname{sech}^{-1}(x) \) by differentiating \( x = \operatorname{sech}(y) \) implicitly. Since \( y \) is a function of \( x \), this requires differentiating both sides of the equation with respect to \( x \), treating \( y \) as an implicit function of \( x \).
The idea here is to differentiate the function \( x = \operatorname{sech}(y) \), which gives us the equation \( 1 = -\operatorname{sech}(y)\tanh(y) \frac{dy}{dx}\). Solve for \( \frac{dy}{dx} \) by isolating it on one side of the equation. This process effectively gives us the rate of change of \( y \) with respect to \( x \), even if \( y \) is not explicitly defined purely in terms of \( x \).
Implicit differentiation is indispensable in situations where equations cannot easily be rearranged to isolate one variable as a function of another.
In this problem, we are tasked with finding the derivative of \( y = \operatorname{sech}^{-1}(x) \) by differentiating \( x = \operatorname{sech}(y) \) implicitly. Since \( y \) is a function of \( x \), this requires differentiating both sides of the equation with respect to \( x \), treating \( y \) as an implicit function of \( x \).
The idea here is to differentiate the function \( x = \operatorname{sech}(y) \), which gives us the equation \( 1 = -\operatorname{sech}(y)\tanh(y) \frac{dy}{dx}\). Solve for \( \frac{dy}{dx} \) by isolating it on one side of the equation. This process effectively gives us the rate of change of \( y \) with respect to \( x \), even if \( y \) is not explicitly defined purely in terms of \( x \).
Implicit differentiation is indispensable in situations where equations cannot easily be rearranged to isolate one variable as a function of another.
Hyperbolic Trigonometric Identities
When working with hyperbolic functions like \( \operatorname{sech}(y) \) and \( \tanh(y) \), it's essential to remember the hyperbolic identities that can simplify these expressions.
One of the key identities used in this exercise is \( \operatorname{sech}^2(y) + \tanh^2(y) = 1 \). This identity is analogous to the Pythagorean identity in trigonometry for sine and cosine functions, and it plays a crucial role in simplifying our derivative expression.
In our context, we also need to express \( \tanh(y) \) in terms of \( x \), because \( x = \operatorname{sech}(y) \). So, \( \tanh(y) = \sqrt{1 - \operatorname{sech}^2(y)} \). Substituting \( x \) for \( \operatorname{sech}(y) \), we find \( \tanh(y) = \sqrt{1 - x^2} \).
Using hyperbolic identities helps transition from a complex-form derivative expression to one that relies solely on \( x \), which is desired when describing derivatives of inverse functions.
One of the key identities used in this exercise is \( \operatorname{sech}^2(y) + \tanh^2(y) = 1 \). This identity is analogous to the Pythagorean identity in trigonometry for sine and cosine functions, and it plays a crucial role in simplifying our derivative expression.
In our context, we also need to express \( \tanh(y) \) in terms of \( x \), because \( x = \operatorname{sech}(y) \). So, \( \tanh(y) = \sqrt{1 - \operatorname{sech}^2(y)} \). Substituting \( x \) for \( \operatorname{sech}(y) \), we find \( \tanh(y) = \sqrt{1 - x^2} \).
Using hyperbolic identities helps transition from a complex-form derivative expression to one that relies solely on \( x \), which is desired when describing derivatives of inverse functions.
Solving Quadratic Equations
Solving quadratic equations is another cornerstone of this exercise, as it comes into play when isolating \( e^y \) during the differentiation process.
Initially, rearranging the expression \( x(e^y + e^{-y}) = 2 \) into a solvable quadratic form of \( xe^{2y} - 2e^y + x = 0 \) allows us to utilize the quadratic formula: \[ e^y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a = x \), \( b = -2 \), and \( c = x \). This process yields the solutions for \( e^y \).
Quadratic equations, particularly those that involve \( e^y \), are central in converting from expressions in terms of \( y \), involving exponential functions, to simplified expressions in terms of \( x \). We pick the solution \( e^y = \frac{1 + \sqrt{1 - x^2}}{x} \) based on the context that \( e^y \) must be positive for real numbers.
Understanding these steps is crucial, as it simplifies the task of finding the derivative in terms of \( x \), leading us toward the final proof.
Initially, rearranging the expression \( x(e^y + e^{-y}) = 2 \) into a solvable quadratic form of \( xe^{2y} - 2e^y + x = 0 \) allows us to utilize the quadratic formula: \[ e^y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a = x \), \( b = -2 \), and \( c = x \). This process yields the solutions for \( e^y \).
Quadratic equations, particularly those that involve \( e^y \), are central in converting from expressions in terms of \( y \), involving exponential functions, to simplified expressions in terms of \( x \). We pick the solution \( e^y = \frac{1 + \sqrt{1 - x^2}}{x} \) based on the context that \( e^y \) must be positive for real numbers.
Understanding these steps is crucial, as it simplifies the task of finding the derivative in terms of \( x \), leading us toward the final proof.
