Problem 39 - University Calculus: Early Transcendentals

Question

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

\(y' = \frac{10x}{x^2 + 1} + \frac{1}{2(1-x)}\)

Step 1: Simplify the Expression Using Logarithmic Properties

Use the properties of logarithms to simplify the given expression. Recall that \( \ln \left( \frac{a}{b} \right) = \ln(a) - \ln(b) \) and \( \ln(a^b) = b \cdot \ln(a) \). Therefore, the expression becomes: \[ y = \ln((x^{2} + 1)^{5}) - \ln(\sqrt{1-x}) \] Applying the power rule of logarithms:\[ y = 5 \ln(x^{2} + 1) - \frac{1}{2} \ln(1-x) \]

Step 2: Differentiate Each Part of the Expression

Differentiate both terms separately with respect to \(x\). For the first term, \(5 \ln(x^2 + 1)\), use the chain rule. For \(\ln(u)\), \(u'\), and the expression becomes:\[ \frac{d}{dx}[5 \ln(x^2 + 1)] = \frac{5 \, (2x)}{x^2 + 1} = \frac{10x}{x^2 + 1} \]For the second term, \(-\frac{1}{2} \ln(1-x)\), again apply the chain rule. The derivative is:\[ \frac{d}{dx}\left[-\frac{1}{2} \ln(1-x)\right] = \frac{-1/2 \, (-1)}{1-x} = \frac{1}{2(1-x)} \]

Step 3: Combine the Derivatives

Combine the derivatives obtained from each term. The derivative of the expression is:\[\frac{d}{dx}y = \frac{10x}{x^2 + 1} + \frac{1}{2(1-x)}\]

Key Concepts

Chain Rule

The chain rule is a powerful tool in calculus for finding the derivative of composite functions. It helps you break down complex derivatives into manageable steps. If you have a function within another function, like \(f(g(x))\), the chain rule says to differentiate the outer function \(f\) while keeping the inner function \(g(x)\) unchanged, then multiply by the derivative of the inner function.
For example, when differentiating \[5 \ln(x^2 + 1)\], we treated \(x^2 + 1\) as the inner function and applied the chain rule:

• Outer function: \(5\ln(u)\) where \(u = x^2 + 1\)
• Derivative of the outer function: \(\frac{5}{u} = \frac{5}{x^2 + 1}\)
• Derivative of the inner function: \(2x\)
• Using the chain rule: \(\frac{d}{dx}[5\ln(x^2 + 1)] = \frac{10x}{x^2 + 1}\)

Breaking it down this way simplifies complex derivatives, making problems easier to solve.

Logarithmic Differentiation

Logarithmic differentiation is helpful when dealing with products or quotients of functions, as well as expressions with powers. It involves applying logarithms to both sides of an equation before differentiating. This not only simplifies your work but also facilitates handling powers, roots, and products more easily.
In our example, we started with:

• \(y = \ln((x^2 + 1)^5) - \ln(\sqrt{1-x})\)
• Differentiating term by term, using logarithmic differentiation simplifies finding derivatives involving functions with exponents or roots.

It gives you a straightforward path to take derivatives of expressions that might otherwise look intimidating. Logarithmic differentiation can save you a lot of effort and reduce mistakes by streamlining the differentiation process.

Logarithmic Properties

Mastering logarithmic properties is essential for effectively simplifying expressions before differentiating them. These properties enable you to break down complex expressions, making them easier to work with.Some fundamental logarithmic properties include:

• \(\ln(\frac{a}{b}) = \ln(a) - \ln(b)\): This property helps to split a logarithm of a quotient into a difference of logarithms.
• \(\ln(a^b) = b \cdot \ln(a)\): This allows you to move the exponent in front of the logarithm, handy for simplifying power products.

In the original exercise, we simplified \(\ln\left(\frac{(x^2 + 1)^5}{\sqrt{1-x}}\right)\) using these properties:

• First, expressed it as a difference: \ \ln((x^2 + 1)^5) - \ln(\sqrt{1-x}) \
• Then, applied the property for powers: \(5 \ln(x^2 + 1) - \frac{1}{2} \ln(1-x)\)

These simplifications made the step-by-step solution clearer and the differentiation process much simpler.