Problem 33 - Thomas Calculus

Question

In Exercises \(5-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\ln \left(\frac{\left(x^{2}+1\right)^{5}}{\sqrt{1-x}}\right) $$

Step-by-Step Solution

Answer

The derivative is \(\frac{10x}{x^2+1} + \frac{1}{2(1-x)}\)."}

Step 1: Use the Logarithmic Property

Apply the logarithmic identity for division: \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\). This gives: \(y = \ln\left((x^2 + 1)^5\right) - \ln\left(\sqrt{1 - x}\right)\).

Step 2: Simplify with Exponent Rules

Use the logarithmic property that \(\ln(a^b) = b \ln(a)\). For the first term, this becomes \(5 \ln(x^2 + 1)\). For the second term, since \(\sqrt{1 - x} = (1 - x)^{1/2}\), it becomes \(\frac{1}{2}\ln(1-x)\). Therefore, \(y = 5 \ln(x^2 + 1) - \frac{1}{2} \ln(1-x)\).

Step 3: Differentiate Each Term

Differentiate \(y = 5 \ln(x^2 + 1) - \frac{1}{2} \ln(1-x)\) with respect to \(x\).The derivative of \(5 \ln(x^2 + 1)\) is: \[5 \cdot \frac{1}{x^2 + 1} \cdot 2x = \frac{10x}{x^2 + 1}\]The derivative of \(-\frac{1}{2} \ln(1-x)\) is: \[-\frac{1}{2} \cdot \frac{1}{1-x} \cdot (-1) = \frac{1}{2(1-x)}\].

Step 4: Combine the Derivatives

Add the derivatives from Step 3 to find the complete derivative:\[\frac{dy}{dx} = \frac{10x}{x^2 + 1} + \frac{1}{2(1-x)}\].

Key Concepts

Logarithmic Differentiation

One of the powerful techniques to find the derivative of complicated functions is logarithmic differentiation. This method can simplify the differentiation process significantly, especially when dealing with products, quotients, or powers of functions.

In the given exercise, we have a complex function inside a logarithm, making it a perfect candidate for logarithmic differentiation. The key steps include:

• Take the natural logarithm of both sides of the equation if needed.
• Use properties of logarithms to expand or simplify the expression.
• Differentiate both sides with respect to the independent variable, using calculated results to find the derivative of the original function.

Properties of logarithms, such as \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \) and \( \ln(a^b) = b \ln(a) \), turn products and powers into manageable terms. Using logarithmic differentiation helps parse the function into simpler components for easier application of other differentiation techniques.

Chain Rule

When differentiating compositions of functions, the chain rule becomes essential. It allows us to differentiate a composite function by setting up a chain of derivatives.

In our step-by-step solution, we employed the chain rule multiple times. For instance, when differentiating \(5 \ln(x^2 + 1)\), we first differentiate the outer function (\(5 \ln\)), giving us \(\frac{1}{x^2 + 1}\). Next, we multiply it by the derivative of the inner function \((x^2 + 1)^\prime = 2x)\). The result is \(5 \cdot \frac{1}{x^2 + 1} \cdot 2x = \frac{10x}{x^2 + 1}\).

This method allows us to dissect a derivative into manageable steps, ensuring we attend to each part of a composite function. It becomes crucial, especially when nested functions appear in complex derivatives.

Product and Quotient Rules

The product and quotient rules are key tools for differentiation, allowing us to handle functions that are products or quotients of two simpler functions.

Although logarithmic differentiation was the main strategy employed, understanding these rules remains vital for calculus. The quotient rule, for example, states: \(\left(\frac{u}{v}\right)^\prime = \frac{u^\prime v - uv^\prime}{v^2}\). This formula applies to functions expressed as one function divided by another.

In the original exercise, the logarithmic properties simplified our function initially without needing to directly apply the quotient rule, but it's necessary to understand how these differentiation techniques intertwine. Using logarithms to convert products and quotients into additions and subtractions of terms is often far simpler and efficient than applying the product/quotient rules directly.