Problem 11 - Calculus for Biology and Medicine

Question

Find \(\frac{d y}{d x}\) if \(\ln \left(x^{2}+y^{2}\right)=3 x y\)

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{3y(x^{2}+y^{2}) - 2x}{2y - 3x(x^{2}+y^{2})} \)

Step 1: Differentiate Both Sides with Respect to x

We have the equation \( \ln \left(x^{2}+y^{2}\right)=3 x y \). To find \( \frac{d y}{d x} \), we will differentiate both sides with respect to \( x \) using implicit differentiation. The derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). So for the left side, we have:\[ \frac{1}{x^{2}+y^{2}} \cdot \left(2x + 2y \cdot \frac{dy}{dx} \right) \]For the right side, use the product rule: \( \frac{d}{dx} (3xy) = 3(x \cdot \frac{dy}{dx} + y) \).

Step 2: Set up the Derivative Equation

After differentiating both sides, we obtain the equation:\[ \frac{1}{x^{2}+y^{2}} \cdot (2x + 2y \cdot \frac{dy}{dx}) = 3 \left( \frac{dy}{dx} \cdot x + y \right) \]

Step 3: Solve for \(\frac{dy}{dx}\)

Multiply through by \( x^{2}+y^{2} \) to clear the fraction:\[ 2x + 2y \cdot \frac{dy}{dx} = 3x \cdot (x^{2}+y^{2}) \cdot \frac{dy}{dx} + 3y \cdot (x^{2}+y^{2}) \]Rearrange the terms to solve for \( \frac{dy}{dx} \):\[ 2y \cdot \frac{dy}{dx} - 3x(x^{2}+y^{2}) \cdot \frac{dy}{dx} = 3y(x^{2}+y^{2}) - 2x \]Bring the \( \frac{dy}{dx} \) terms to one side:\[ \frac{dy}{dx} \left( 2y - 3x(x^{2}+y^{2}) \right) = 3y(x^{2}+y^{2}) - 2x \]And solve for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{3y(x^{2}+y^{2}) - 2x}{2y - 3x(x^{2}+y^{2})} \]

Key Concepts

Understanding Derivatives

A derivative is a fundamental concept in calculus. It measures how a function changes as its input changes. When we talk about finding the derivative, represented by \( \frac{d y}{d x} \), we are essentially looking at the rate of change of one variable with respect to another. In this exercise, we are applying implicit differentiation, which is used when a function is not given explicitly.In simpler terms, implicit differentiation allows us to find derivatives for equations where \( y \) is not isolated on one side. It's a handy technique when dealing with equations involving multiple variables like \( \ln(x^2 + y^2) = 3xy \). By differentiating both sides with respect to \( x \), you track how changes in \( x \) affect changes in \( y \).

Using the Product Rule

The product rule is a tool used to differentiate expressions like \( u \cdot v \), where both \( u \) and \( v \) are functions of \( x \). The rule states: \[ \frac{d}{dx}(u \, v) = u' \, v + u \, v' \]In our problem, you apply the product rule to the right side of the equation \( 3xy \). Here, \( 3 \) is a constant factor, so it comes out of the derivative. Then you assign \( u = x \) and \( v = y \) and find their derivatives:

• \( u' = 1 \) since the derivative of \( x \) with respect to \( x \) is 1.
• \( v' = \frac{d y}{d x} \), which represents the change in \( y \) with respect to \( x \).

Putting it together, the derivative of \( 3xy \) is \( 3(x \cdot v' + v \cdot 1) \). This simplifies to \( 3(x \cdot \frac{d y}{d x} + y) \), showing how both \( x \) and \( y \) interact as \( x \) changes.

Applying Logarithmic Differentiation

Logarithmic differentiation is especially useful when dealing with products or quotients of functions, or powers of functions. It simplifies differentiation by taking the natural log (\( \ln \)) of both sides of an equation. This approach is ideal when your function is complex, as seen in the exercise.For \( \ln(x^2 + y^2) = 3xy \), you differentiate each side separately. For the left side, notice how the derivative of a natural log function \( \ln(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). This splits the differentiation into easier parts:

• Identify \( u = x^2 + y^2 \).
• Then differentiate inside \( u \), resulting in \( 2x + 2y \frac{dy}{dx} \).

Thus, the left side becomes \( \frac{1}{x^2 + y^2}(2x + 2y \frac{dy}{dx}) \). Logarithmic differentiation lets you maintain the structure of the original equation while isolating \( \frac{dy}{dx} \) for calculation.