Problem 63 - Essential Calculus Early Transcendentals

Question

Find \(y^{\prime}\) if \(y=\ln \left(x^{2}+y^{2}\right)\)

Step-by-Step Solution

Answer

\(y^{\prime} = \frac{2x}{x^2 + y^2 - 2y}\)."

Step 1: Differentiate with respect to x

We start by differentiating both sides of the equation with respect to \(x\). On the left side, the derivative is \(y^{\prime}\). On the right side, use the chain rule. The derivative of \(\ln(u)\) is \(\frac{1}{u} \frac{du}{dx}\) where \( u = x^2 + y^2 \).

Step 2: Apply chain rule to the right side

First, identify \( u = x^2 + y^2 \). Then, \( \frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx} \). Next, find \( \frac{du}{dx} = \frac{d}{dx}(x^2 + y^2) \).

Step 3: Differentiate u with respect to x

Calculate \( \frac{du}{dx} \). The derivative of \(x^2\) is \(2x\) and the derivative of \(y^2\) with respect to \(x\) is \(2yy^{\prime}\) using implicit differentiation. So \( \frac{du}{dx} = 2x + 2yy^{\prime} \).

Step 4: Substitute back into chain rule result

Substitute \( \frac{du}{dx} = 2x + 2yy^{\prime} \) into equation from Step 2: \( \frac{1}{x^2 + y^2}(2x + 2yy^{\prime}) \).

Step 5: Set equations equal and solve for y-prime

We have \(y^{\prime} = \frac{1}{x^2 + y^2}(2x + 2yy^{\prime})\). Multiply both sides by \(x^2 + y^2\) to eliminate the fraction: \((x^2 + y^2)y^{\prime} = 2x + 2yy^{\prime}\).

Step 6: Isolate y-prime

Rearrange to have terms with \(y^{\prime}\) on one side: \((x^2 + y^2)y^{\prime} - 2yy^{\prime} = 2x\). Factor \(y^{\prime}\): \(y^{\prime}(x^2 + y^2 - 2y) = 2x\).

Step 7: Solve for y-prime

Isolate \(y^{\prime}\) by dividing both sides by \(x^2 + y^2 - 2y\): \(y^{\prime} = \frac{2x}{x^2 + y^2 - 2y}\).

Key Concepts

Chain Rule

The chain rule is a crucial concept in calculus used to differentiate composite functions. When faced with a function like \( \ln(x^2 + y^2) \), the chain rule helps us tackle the derivative efficiently. The formula for the chain rule states that

• if \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

Here, our outer function is the natural logarithm, \( \ln(u) \), and our inner function is \( u = x^2 + y^2 \). The derivative of a natural logarithm is \( \frac{1}{u} \), and the chain rule dictates that we also need the derivative of the inner function \( u \) with respect to \( x \).
To apply the chain rule correctly, it is important to:

• First, identify the inner and the outer functions.
• Then, find the derivative of the outer function while treating the inner function as a single variable.
• Finally, multiply this by the derivative of the inner function with respect to \( x \).

Implicit Differentiation

Implicit differentiation is an advanced technique used when differentiating equations where variables are intermixed, as with equations like \( y = \ln(x^2 + y^2) \). Here, \( y \) is a function of \( x \), but it is not solved explicitly.
This approach involves:

• Treating \( y \) as an implicit function of \( x \).
• Applying standard differentiation rules, recognizing that \( \frac{dy}{dx} = y' \).

When differentiating terms involving \( y \), like \( y^2 \), with respect to \( x \), use the chain rule to yield \( 2yy' \), where \( y' \) is \( \frac{dy}{dx} \).
Implicit differentiation allows us to solve for \( y' \) even when it cannot be isolated easily, making it a powerful tool in calculus.

Natural Logarithm Derivatives

The derivative of the natural logarithm function, \( \ln(u) \), is straightforward but vital to master. This derivative is expressed as:

• \( \frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx} \)

In our example, \( u = x^2 + y^2 \), which requires us to employ the chain rule to find \( \frac{du}{dx} \). Understanding that \( \frac{1}{u} \) is part of this derivative is crucial for solving logarithmic differentiation problems accurately.
Pay attention to:

• Identifying the inner function \( u \).
• Applying the chain rule diligently.

This method is particularly beneficial in cases involving complex functions inside the logarithm, ensuring precision even when variables are entangled, as seen in implicit differentiation scenarios.