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Magazine Chapter 7: Problem 67
Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=\sqrt[3]{\frac{x(x-2)}{x^{2}+1}}$$
Short Answer
Expert verified
\(\frac{dy}{dx} = \sqrt[3]{\frac{x(x-2)}{x^2 + 1}} \cdot \frac{1}{3}\left(\frac{2(x-1)}{x(x-2)} - \frac{2x}{x^2+1}\right)\)
Step by step solution
01
Apply Logarithm to Both Sides
Start by taking the natural logarithm of both sides of the equation. Take the natural log of both sides: \( rac{1}{3} \ln\left(y\right) = \ln\left(\sqrt[3]{\frac{x(x-2)}{x^{2}+1}}\right)\). Now, using the properties of logarithms, this simplifies to \(\ln(y) = \frac{1}{3}\left(\ln\left(x(x-2)\right) - \ln\left(x^2 + 1\right)\right)\).
02
Differentiate Both Sides Implicitly
Differentiate both sides with respect to \(x\). For the left side, use the chain rule: \(\frac{d}{dx}[\ln(y)] = \frac{1}{y}\frac{dy}{dx}\). For the right side, differentiate using the chain rule and the fact that \(\frac{d}{dx}[\ln(u)] = \frac{1}{u}\frac{du}{dx}\): \(\frac{1}{3}\left(\frac{d}{dx}[\ln(x(x-2))] - \frac{d}{dx}[\ln(x^2 + 1))\right)\).
03
Differentiate Logarithmic Terms
For \(\ln(x(x-2))\), let \(u = x(x-2)\), then \(u' = (x-2) + x = 2x - 2\), so \(\frac{d}{dx}[\ln(x(x-2))] = \frac{1}{x(x-2)}(2x - 2)\). For \(\ln(x^2 + 1)\), let \(v = x^2 + 1\), then \(v' = 2x\), so \(\frac{d}{dx}[\ln(x^2 + 1)] = \frac{1}{x^2 + 1}(2x)\).
04
Simplify Derivative Expression
Plug the differentiated terms back into our equation: \(\frac{1}{y}\frac{dy}{dx} = \frac{1}{3}\left(\frac{2x-2}{x(x-2)} - \frac{2x}{x^2+1}\right)\). Simplify: \(\frac{1}{y}\frac{dy}{dx} = \frac{1}{3}\left(\frac{2(x-1)}{x(x-2)} - \frac{2x}{x^2+1}\right)\).
05
Solve for \(\frac{dy}{dx}\)
Multiply both sides by \(y\) to solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = y \cdot \frac{1}{3}\left(\frac{2(x-1)}{x(x-2)} - \frac{2x}{x^2+1}\right)\). Substitute the example value of \(y\) back in: \(y = \sqrt[3]{\frac{x(x-2)}{x^2 + 1}}\). Thus: \(\frac{dy}{dx} = \sqrt[3]{\frac{x(x-2)}{x^2 + 1}} \cdot \frac{1}{3}\left(\frac{2(x-1)}{x(x-2)} - \frac{2x}{x^2+1}\right)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Natural Logarithm
The natural logarithm, denoted as \( \ln(x) \), is a logarithmic function with the base \( e \), where \( e \) is an irrational number approximately equal to 2.71828. Unlike other logarithms, the natural logarithm has special properties that make it particularly useful in calculus and mathematical analysis.
Properties of Natural Logarithm:
Properties of Natural Logarithm:
- Simplification by log rules: It allows us to simplify complex expressions through the use of its properties. For example, \( \ln(a \cdot b) = \ln(a) + \ln(b) \) and \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \).
- Function inversion: The natural logarithm is the inverse of the exponential function \( e^x \), so \( e^{\ln(x)} = x \) and \( \ln(e^x) = x \).
- Derivatives: The derivative of \( \ln(x) \) is straightforward, \( \frac{d}{dx}[\ln(x)] = \frac{1}{x} \), which is very helpful when differentiating.
Chain Rule
In calculus, the chain rule is a fundamental technique for differentiating compositions of functions. It allows us to differentiate a composite function by breaking it down into its constituent parts. The chain rule states that if a function \( y = f(g(x)) \), the derivative \( \frac{dy}{dx} \) is:\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x)\]This means you differentiate the outer function evaluated at the inner function and multiply it by the derivative of the inner function.
Applying the Chain Rule:
Applying the Chain Rule:
- Identify the inner and outer functions: Recognize the nested functions and label them. For example, in \( \ln(u) \), \( u \) is the inner function.
- Differentiation: Use the rule to sequentially differentiate. Identify \( u \) and differentiate it to find \( u' \) as shown in the example solution where \( u = x(x-2) \).
- Combine results: Multiply the derivative of the outer function by the derivative of the inner function. For example, \( \frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot u' \).
Implicit Differentiation
Implicit differentiation is a way to find the derivative when the function is not explicitly solved for one of the variables. Instead of solving for \( y \) in terms of \( x \) first, implicit differentiation allows us to differentiate directly, treating \( y \) as an implicit function of \( x \).
Steps in Implicit Differentiation:
Steps in Implicit Differentiation:
- Differentiation of both sides: Start by differentiating both sides of the given equation with respect to \( x \), treating all occurrences of \( y \) as functions of \( x \).
- Application of chain rule: Each time you differentiate \( y \) with respect to \( x \), apply the chain rule, which involves multiplying by \( \frac{dy}{dx} \) (because \( \frac{d}{dx}[y] = \frac{dy}{dx} \)).
- Solve for \( \frac{dy}{dx} \): Once all terms involving \( \frac{dy}{dx} \) are isolated, solve the resulting equation for \( \frac{dy}{dx} \). This is often the step where simplification happens.
