Problem 56 - Thomas Calculus

Question

In Exercises \(55-68,\) use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$ y=\sqrt{\left(x^{2}+1\right)(x-1)^{2}} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = \sqrt{(x^2 + 1)(x - 1)^2} \left( \frac{x}{x^2 + 1} + \frac{1}{x - 1} \right) \).

Step 1: Rewrite the Function using Logarithms

Start by rewriting the function in a form that is easier to manipulate with logarithms. Given \[ y = \sqrt{(x^2 + 1)(x - 1)^2}, \]you can express this as \[ y = [(x^2 + 1)(x - 1)^2]^{\frac{1}{2}}. \] Taking the natural logarithm of both sides, we have: \[ \ln(y) = \frac{1}{2} \ln((x^2 + 1)(x - 1)^2). \] Utilize the logarithmic identity \( \ln(ab) = \ln(a) + \ln(b) \) to further simplify:\[ \ln(y) = \frac{1}{2} \left( \ln(x^2 + 1) + \ln((x - 1)^2) \right). \] Further simplify using the power rule for logarithms, \( \ln(a^b) = b\ln(a) \):\[ \ln(y) = \frac{1}{2} \ln(x^2 + 1) + \ln(x - 1). \]

Step 2: Differentiate Implicitly with Respect to x

Differentiate both sides of the log-transformed equation implicitly with respect to \(x\):For the left-hand side, the derivative of \(\ln(y)\) with respect to \(x\) is \(\frac{1}{y} \frac{dy}{dx}\).For the right-hand side, apply the chain and product rules:- The derivative of \(\frac{1}{2} \ln(x^2 + 1)\) is \(\frac{1}{2} \cdot \frac{1}{x^2 + 1} \cdot 2x = \frac{x}{x^2 + 1}\).- The derivative of \(\ln(x - 1)\) is \(\frac{1}{x - 1}\).Putting it together, we have:\[ \frac{1}{y} \frac{dy}{dx} = \frac{x}{x^2 + 1} + \frac{1}{x - 1}. \]

Step 3: Solve for dy/dx

Isolate \(\frac{dy}{dx}\) on the left side by multiplying both sides by \(y\):\[ \frac{dy}{dx} = y \left( \frac{x}{x^2 + 1} + \frac{1}{x - 1} \right). \]Re-substitute the original expression for \(y\):\[ y = \sqrt{(x^2 + 1)(x - 1)^2}, \]thus,\[ \frac{dy}{dx} = \sqrt{(x^2 + 1)(x - 1)^2} \left( \frac{x}{x^2 + 1} + \frac{1}{x - 1} \right). \]

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a function changes at any given point. The derivative of a function at a certain point measures the slope of the tangent line to the function's graph at that point.
For example, if we have a function \( f(x) \), its derivative is written as \( f'(x) \) or \( \frac{df}{dx} \). This tells us how the function \( f \) changes with respect to changes in \( x \).

• The notation \( \frac{dy}{dx} \) is often used, where \( y \) is a function of \( x \).
• It's an essential tool for finding rates of change, such as speed, growth, and other dynamic processes.

To find a derivative, one can use several rules and techniques, such as the power rule, product rule, quotient rule, and chain rule. In the case of complex functions, logarithmic differentiation is a handy method to simplify the differentiation process, as showcased in our original exercise.

Exploring Implicit Differentiation

Implicit differentiation is a technique used when a function is not easily solvable for one of the variables. This approach allows us to find the derivative of functions that are not clearly defined or when variables cannot be easily separated.
In traditional explicit differentiation, you can directly solve for the function in terms of one variable, say \( y = f(x) \). However, with implicit differentiation, both variables are interwoven, requiring a different method.

• We differentiate both sides of an equation with respect to \( x \).
• Whenever you differentiate a term containing \( y \), use the chain rule to include \( \frac{dy}{dx} \).

In the context of our solution, we used implicit differentiation after expressing the function in logarithmic form. This allowed us to systematically take derivatives of each component, properly handling the implicit relationship between \( y \) and \( x \).

The Role of the Chain Rule

The chain rule is a critical technique for differentiating composite functions, where one function is nested inside another. It helps you work through these layers by differentiating step by step, from the outer to the inner functions.
In simple terms, if you have a composite function \( g(f(x)) \), the chain rule states that the derivative can be calculated as: \[ (g(f(x)))' = g'(f(x)) \cdot f'(x). \]

• This means you take the derivative of the outer function, keeping the inner function the same, and multiply it by the derivative of the inner function.
• This is particularly useful when functions are nested or intertwined, as it simplifies complex differentiation problems.

In our exercise, the chain rule facilitated differentiation after transforming the function through logarithmic differentiation, ensuring each part was accurately addressed.