Question

Find \(\frac{d y}{d x}\) using implicit differentiation. $$\ln \left(x^{2}+x y+y^{2}\right)=1$$

Short answer

\( \frac{dy}{dx} = \frac{-2x - y}{x + 2y} \)

Step-by-step solution

Differentiate Both Sides with Respect to x

To find \( \frac{dy}{dx} \) using implicit differentiation, start by differentiating both sides of the equation with respect to \( x \). The left side involves a natural logarithm, which differentiates to \( \frac{1}{x^2 + xy + y^2} \cdot \left(\frac{d}{dx}(x^2 + xy + y^2)\right) \). The right side differentiates to 0 because it is constant.

Differentiate Inside the Logarithm

Continue differentiating \( x^2 + xy + y^2 \) with respect to \( x \). The derivative of \( x^2 \) is \( 2x \). For \( xy \), use the product rule: \( x \cdot \frac{dy}{dx} + y \cdot 1 = x \frac{dy}{dx} + y \). The derivative of \( y^2 \) is \( 2y \cdot \frac{dy}{dx} \) using the chain rule. Combine these results to get: \( 2x + (x + 2y)\frac{dy}{dx} + y \).

Substitute Back into the Derivative

Substitute the differentiated expression back into the context of Step 1's result: \[ \frac{1}{x^2 + xy + y^2} \left( 2x + (x + 2y)\frac{dy}{dx} + y \right) = 0 \].

Simplify and Solve for \( \frac{dy}{dx} \)

To solve for \( \frac{dy}{dx} \), multiply both sides by \( x^2 + xy + y^2 \) to remove the denominator, resulting in: \[ 2x + (x + 2y)\frac{dy}{dx} + y = 0 \].Isolate \( \frac{dy}{dx} \): \[ (x + 2y)\frac{dy}{dx} = -2x - y \].Finally, divide by \( x + 2y \) to obtain:\[ \frac{dy}{dx} = \frac{-2x - y}{x + 2y} \].

Key concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a logarithm to the base \( e \), where \( e \approx 2.71828 \). In calculus, the natural logarithm is often encountered when differentiating products or powers that involve the constant \( e \). The key property of the natural logarithm that is useful in differentiation is its derivative.
The derivative of \( \ln(u) \) with respect to \( x \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). This highlights not only the chain rule but also how crucial it is to first understand what the argument of the logarithm represents.
In our exercise, \( \ln(x^2 + xy + y^2) = 1 \), the natural logarithm of a complex expression requires careful differentiation of both the logarithm itself and its argument.

Product Rule

The product rule is a fundamental differentiation rule used when taking the derivative of a product of two functions. If you have a product \( u(x) \cdot v(x) \), the product rule states:

• \( \frac{d}{dx}[u \cdot v] = u' \cdot v + u \cdot v' \)

This rule allows the derivative to be expanded into a sum of two terms, each involving both functions and their derivatives.
In the given problem, the product rule is applied to the term \( xy \) within the logarithmic argument. Here, \( x \) and \( y \) are treated as separate functions. The derivative becomes \( x \frac{dy}{dx} + y \) because \( x \) is constant with respect to \( y \), and \( y \) is a variable dependent on \( x \). This application helps to find the correct expression inside the logarithm's derivative.

Chain Rule

The chain rule is vital when differentiating compositions of functions. It allows us to differentiate the outer function while considering the derivative of the inner function. If \( y = f(g(x)) \), the chain rule state:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

In implicit differentiation, we often need to apply the chain rule more than once, particularly when dealing with expressions involving multiple variables.
For example, when differentiating the term \( y^2 \), the chain rule leads to the term \( 2y \cdot \frac{dy}{dx} \). This emphasizes not just the derivative of the squared terms, but also the fact that \( y \), being dependent on \( x \), requires the inclusion of \( \frac{dy}{dx} \) as per the chain rule. The chain rule allows the complexity of nested functions to be unraveled, making differentiation manageable.

Derivative Calculation

Derivative calculation is the process of finding the rate at which a function changes with respect to a variable. This can become complex when dealing with implicit functions, where both variables influence each other. Steps to take include:

• Differentiate all terms with respect to \( x \).
• Apply the product rule and chain rule where necessary.
• Substitute and rearrange terms to isolate \( \frac{dy}{dx} \).

In this task, you start by isolating \( \frac{dy}{dx} \) in the expression. By combining like terms and rearranging, you solve for \( \frac{dy}{dx} \). The derivative simplifies to \[ \frac{dy}{dx} = \frac{-2x - y}{x + 2y} \]. This process underscores the precision needed in handling terms and manipulations in the context of implicit differentiation.