Question

If \(x=\left[\left(e^{2 y}+1\right) /\left(e^{2 y}-1\right)\right]\) then \((d y / d x)=\) (a) \(\left[1 /\left(1+\mathrm{x}^{2}\right)\right]\) (b) \(\left[1 /\left(\mathrm{x}^{2}-1\right)\right]\) (c) \(\left[1 /\left(1-x^{2}\right)\right]\) (d) \(\left[1 /\left(x^{2}-1\right)\right]\)

Short answer

The short answer is \(\frac{dy}{dx} = \frac{1}{x^2-1}\).

Step-by-step solution

Finding y(x)

To find the inverse function, which means finding y as a function of x, we must rewrite x as follows: \(x = \frac{e^{2y}+1}{e^{2y}-1}\) We can rewrite this equation as: \(x(e^{2y}-1) = e^{2y}+1\) Now, isolate e^(2y): \(xe^{2y} - e^{2y} = 1 + x\) Factor out e^(2y): \(e^{2y}(x-1) = x+1\) Now, find e^(2y): \(e^{2y} = \frac{x+1}{x-1}\) Now, apply the natural logarithm to both sides: \(2y = \ln{\frac{x+1}{x-1}}\) Finally, find y as a function of x: \(y = \frac{1}{2}\cdot \ln{\frac{x+1}{x-1}}\)

Finding the derivative dy/dx

Now, we can find the derivative of y with respect to x, using the chain rule: \(\frac{dy}{dx} = \frac{1}{2}\cdot\frac{d}{dx}\left(\ln{\frac{x+1}{x-1}}\right)\) Differentiate the natural logarithm function: \(\frac{dy}{dx} = \frac{1}{2}\cdot\frac{1}{\frac{x+1}{x-1}}\cdot\frac{d}{dx}\left(\frac{x+1}{x-1}\right)\) Now, differentiate the fraction using quotient rule: \(\frac{d}{dx}(\frac{x+1}{x-1}) = \frac{(x-1)(1)-(x+1)(1)}{(x-1)^2}\) \(\frac{d}{dx}(\frac{x+1}{x-1}) = \frac{-2}{(x-1)^2}\) Now, substitute back into the equation of dy/dx: \(\frac{dy}{dx} = \frac{1}{2}\cdot\frac{1}{\frac{x+1}{x-1}}\cdot\frac{-2}{(x-1)^2}\) Simplify the equation: \(\frac{dy}{dx} = \frac{-1}{\frac{(x+1)(x-1)^2}{x-1}}\) \(\frac{dy}{dx} = \frac{-1}{(x+1)(x-1)}\)

Comparing the result with the answer choices

Our calculated dy/dx is given by: \(\frac{dy}{dx} = \frac{-1}{(x+1)(x-1)}\) This result matches option (b), which is: \(\frac{dy}{dx} = \left[\frac{1}{x^2-1}\right]\) So, the correct answer is (b).

Key concepts

Inverse Functions

Inverse functions are like two sides of a coin in mathematics—each side uniquely matching the other. To find an inverse function, we interchange the roles of the dependent and independent variables. This means expressing the original independent variable, often x, in terms of the dependent variable, usually y, and then solving for y as a standalone function.

Doing this essentially reverses the action of the original function. If the function takes you from A to B, then the inverse takes you back from B to A. In the context of the exercise, when given the function for x in terms of y, finding its inverse involves isolating y and making it the subject of the equation. This way, students can see the 'backwards journey' from x to y.

Natural Logarithm

The natural logarithm, denoted as \(ln\), is a special kind of log that has the mathematical constant e (approximately 2.718) as its base. The natural logarithm of any positive number x is defined as the power to which e has to be raised to yield that number. When the base e is raised to this power, you get x back.

In our exercise, when we have \(e^{2y}\), taking the natural logarithm of both sides gives us a tool to 'unpack' the exponent and bring the y down to earth where we can work with it. Understanding the relationship between exponentials and logarithms is crucial for tackling problems involving growth and decay, among many other real-world applications.

Chain Rule

The chain rule in differential calculus is a method for finding the derivative of a composite function. Think of it as a way to peel back layers of a function. It's saying, 'To find the rate at which the whole system changes, first understand how the outer part changes with the inner part, then how the inner part changes on its own, and multiply those rates.'

In practice, if a function y is formed by the composition of two functions, g and h, where \(y = g(h(x))\), then the derivative of y with respect to x is \(g'(h(x)) \cdot h'(x)\). The exercise implements the chain rule to differentiate the natural logarithm of a quotient, where the inner part is the quotient and the outer part is the logarithm function.

Quotient Rule

The quotient rule is another integral part of calculus, used when you need to find the derivative of a ratio of two functions. The rule states that for two differentiable functions, f(x) and g(x), the derivative of their quotient is given by:
\br>\( \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{[g(x)]^2} \).

In simpler terms, it's like saying 'to find the rate of change of a fraction, take the bottom times the derivative of the top, subtract the top times the derivative of the bottom, and divide it all by the bottom squared.' This rule is used in the exercise to find the derivative of the expression inside the logarithm. Such foundational rules are incredibly efficient for managing even the most complex of fractions.