Problem 25 - Thomas Calculus

Question

Use implicit differentiation to find \(d y / d x\) in Exercises \(19-32\) $$ y^{2}=\frac{x-1}{x+1} $$

Step-by-Step Solution

Answer

The solution is \(\frac{dy}{dx} = \frac{1}{y(x+1)^2}\).

Step 1: Differentiate both sides implicitly

Start with the equation \(y^2 = \frac{x-1}{x+1}\). Differentiate both sides with respect to \(x\). For the left side, use the chain rule. We have \(\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}\). For the right side, use the quotient rule: \[\frac{d}{dx}\left(\frac{x-1}{x+1}\right) = \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2} = \frac{x+1-x+1}{(x+1)^2} = \frac{2}{(x+1)^2}.\] Hence, differentiate both sides to get \(2y \frac{dy}{dx} = \frac{2}{(x+1)^2}\).

Step 2: Solve for \(\frac{dy}{dx}\)

Isolate \(\frac{dy}{dx}\) by dividing both sides of the equation by \(2y\). This gives \(\frac{dy}{dx} = \frac{2}{(x+1)^2} \times \frac{1}{2y}\). Simplify this to \[\frac{dy}{dx} = \frac{1}{y(x+1)^2}.\]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus. It is used when differentiating compositions of functions. Essentially, the chain rule helps us find the derivative of a function that is nested within another function.
For instance, when differentiating a function like \( y^2 \), we recognize that \( y \) is a function of \( x \). Therefore, instead of just dealing with \( y \), the chain rule comes into play, and we must multiply the derivative of \( y^2 \) with the derivative of \( y \) with respect to \( x \).
This process can be seen when finding \( \frac{d}{dx}(y^2) \). By applying the chain rule, it becomes \( 2y \frac{dy}{dx} \). Here, the derivative of \( y^2 \) is \( 2y \), and then we multiply by \( \frac{dy}{dx} \). This is the essence of the chain rule, a technique frequently used in differential calculus.

Quotient Rule

The quotient rule is another critical tool used when we need to differentiate a division of two functions. When you have a function \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the quotient rule becomes your go-to method.
The formula for the quotient rule is: \[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\]It shows that you'll need to find the derivatives of both the numerator and denominator separately. Then, carefully apply the formula.
In the given exercise, we applied the quotient rule to differentiate \( \frac{x-1}{x+1} \). By following the steps, we subtract the product of the derivative of the numerator and the denominator derivative and divide by the square of the denominator. This yields a simplified result of \( \frac{2}{(x+1)^2} \).
The quotient rule is particularly useful in scenarios where complex fractions are involved.

Derivatives

Derivatives are at the core of calculus as they represent rates of change. Understanding how to find derivatives of functions is essential for solving problems related to motion, growth, and many other real-world scenarios.
In simple terms, the derivative of a function at a particular point tells us the slope of the tangent line to the function at that point. It's the mathematical way to capture how one quantity changes in relation to another.
When applied, as in this exercise, finding \( dy/dx \) implies identifying how \( y \) changes with respect to changes in \( x \). By using rules of differentiation like the chain rule and quotient rule, we can systematically approach and solve these problems.
Grasping the concept of derivatives unlocks the door to a deeper understanding of many dynamic systems, making it a vital skill in both mathematics and applied fields.

Implicit Differentiation Process

Implicit differentiation is a powerful technique for finding derivatives, especially when it is difficult or impossible to solve for one variable in terms of another explicitly. In such cases, you need to differentiate both sides of an equation with respect to \( x \) and then solve for \( \frac{dy}{dx} \).
The process often involves using other differentiation rules like the chain rule to manage variables that are intertwined within the equation.

• Start by differentiating both sides of the equation with respect to \( x \).
• Apply the chain or quotient rules where necessary for complex expressions.
• Once differentiation is complete, solve for \( \frac{dy}{dx} \) by isolating it on one side of the equation.

This technique is illustrated in our exercise by differentiating the equation \( y^2 = \frac{x-1}{x+1} \). By differentiating implicitly, we first transform and simplify the components using our differentiation rules, ultimately isolating \( \frac{dy}{dx} \) to find our solution. Implicit differentiation enables us to tackle problems that do not lend themselves to easy algebraic manipulations.