Question

In Exercises \(1-8,\) find \(d y / d x\). $$y^{2}=\frac{x-1}{x+1}$$

Short answer

\(\frac{dy}{dx} = \frac{1}{y(x+1)^{2}}\)

Step-by-step solution

Apply Chain Rule to differentiate the left hand side

Differentiate \(y^{2}\) with respect to \(x\) using Chain Rule. The derivative of \(y^{2}\) with respect to \(y\) is \(2y\) and the derivative of \(y\) with respect to \(x\) is \( \frac{dy}{dx}\). The derivative \( \frac{dy^{2}}{dx} \) is therefore \(2y \frac{dy}{dx}\).

Apply the Quotient Rule to differentiate the right hand side

Differentiate \(\frac{x-1}{x+1}\) with respect to \(x\). By the rule for differentiation of quotients (often called the Quotient Rule), if \( u(x) = \frac{f(x)}{g(x)} \), where \(f(x)\) and \(g(x)\) are differentiable functions, then \( u'(x) = \frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^{2}} \). Consequently, the derivative of \(\frac{x-1}{x+1}\) is \(\frac{(1)(x+1)-(1)(x-1)}{(x+1)^{2}}\), which simplifies to \(\frac{2}{(x+1)^{2}}\).

Equate derivatives to solve for dy/dx

Since \(2y \frac{dy}{dx}\) is equal to \(\frac{2}{(x+1)^{2}}\), the derivative \( \frac{dy}{dx} \) can be given by the formula \(\frac{dy}{dx} = \frac{1}{y(x+1)^{2}}\).

Key concepts

Derivative of a Function

The derivative of a function at a certain point is a fundamental concept in calculus that represents the rate at which the function is changing at that particular point. It is often described as the 'instantaneous rate of change' or the 'slope' of the function's graph.

For example, if you have a function that describes the distance traveled over time, its derivative will give you the velocity at any moment, which is how fast the distance is changing over time. In the given exercise, finding the derivative, or \(d y / d x\), is essentially the process of determining how the variable \(y\) changes with respect to \(x\).

When functions are more complex than simple polynomials, special rules such as the Chain Rule and Quotient Rule come into play. These rules are algorithms for differentiating combinations of functions, crucial for handling a variety of expressions one may encounter in calculus.

Chain Rule

The Chain Rule is a formula to compute the derivative of a composite function. When you have a function within another function, the derivative of this composite function is not straightforward. This is where the Chain Rule comes in handy.

Mathematically, if \( f \circ g\) represents the composite function where \( f\) is applied to the result of \( g\), and you wish to find the derivative of this composite function with respect to \(x\), the Chain Rule tells us to first take the derivative of the outer function \(f\) with respect to its argument (usually called \(u\)) and multiply it by the derivative of the inner function \(g\) with respect to \(x\). The formula is given by \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \).

In the context of the provided exercise, the Chain Rule is utilized to find the derivative of \(y^2\) with respect to \(x\). Since \(y\) is itself a function of \(x\), you multiply the derivative of the outer function, \(2y\), by the derivative of the inner function, \(dy/dx\), giving you \(2y \frac{dy}{dx}\).

Quotient Rule

The Quotient Rule is a technique for taking the derivative of a quotient of two functions. It addresses the question of how to find the rate of change of a ratio where both the numerator and the denominator are also changing.

The Quotient Rule states that for two differentiable functions \(u(x)\) and \(v(x)\), the derivative of their quotient \(u(x)/v(x)\) is given by: \[ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - v'(x)u(x)}{[v(x)]^2} \.

\] This is illustrated in the original problem where the derivative of the right-hand side, \(\frac{x-1}{x+1}\), is found. The numerator and the denominator are both functions of \(x\), and their derivatives are \(1\) and \(0\) respectively. According to the Quotient Rule, the derivative thus becomes \( \frac{(1)(x+1)-(1)(x-1)}{(x+1)^2}\), which simplifies to \( \frac{2}{(x+1)^2}\).

This rule is crucial when dealing with fractions in differentiation, and understanding it is crucial to solve problems effectively where division of functions occurs. It is also important to note that both functions \(u(x)\) and \(v(x)\) must be differentiable for this rule to apply.