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Magazine Chapter 3: Problem 63
Find \(y^{\prime}\) if \(y=\ln \left(x^{2}+y^{2}\right)\)
Short Answer
Expert verified
\(y^{\prime} = \frac{2x}{x^2 + y^2 - 2y}\)."
Step by step solution
01
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 \( u = x^2 + y^2 \).
02
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 \( \frac{du}{dx} = \frac{d}{dx}(x^2 + y^2) \).
03
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 \( \frac{du}{dx} = 2x + 2yy^{\prime} \).
04
Substitute back into chain rule result
Substitute \( \frac{du}{dx} = 2x + 2yy^{\prime} \) into equation from Step 2: \( \frac{1}{x^2 + y^2}(2x + 2yy^{\prime}) \).
05
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: \((x^2 + y^2)y^{\prime} = 2x + 2yy^{\prime}\).
06
Isolate y-prime
Rearrange to have terms with \(y^{\prime}\) on one side: \((x^2 + y^2)y^{\prime} - 2yy^{\prime} = 2x\). Factor \(y^{\prime}\): \(y^{\prime}(x^2 + y^2 - 2y) = 2x\).
07
Solve for y-prime
Isolate \(y^{\prime}\) by dividing both sides by \(x^2 + y^2 - 2y\): \(y^{\prime} = \frac{2x}{x^2 + y^2 - 2y}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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
To apply the chain rule correctly, it is important to:
- if \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
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:
Implicit differentiation allows us to solve for \( y' \) even when it cannot be isolated easily, making it a powerful tool in calculus.
This approach involves:
- Treating \( y \) as an implicit function of \( x \).
- Applying standard differentiation rules, recognizing that \( \frac{dy}{dx} = y' \).
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:
Pay attention to:
- \( \frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx} \)
Pay attention to:
- Identifying the inner function \( u \).
- Applying the chain rule diligently.
