Question

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

Short answer

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

Step-by-step solution

Differentiate Both Sides with Respect to x

We start by taking the derivative of both sides of the equation with respect to \(x\). The equation we have is \(\ln(x^2 + y^2) = e\). The derivative of a constant \(e\) is 0. So, we need to focus on the left side, which involves the chain rule: the derivative of \(\ln(u)\) is \(\frac{1}{u} \cdot \frac{du}{dx}\).

Apply the Chain Rule to the Left Side

For \(\ln(x^2 + y^2)\), let \(u = x^2 + y^2\). We have:\[\frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx}\]First, compute \(\frac{du}{dx}\) where \(u = x^2 + y^2\). This requires differentiating \(x^2\) and \(y^2\) with respect to \(x\). - The derivative of \(x^2\) with respect to \(x\) is \(2x\).- The derivative of \(y^2\) with respect to \(x\) is \(2y \cdot \frac{dy}{dx}\) due to implicit differentiation.Thus, \(\frac{du}{dx} = 2x + 2y \cdot \frac{dy}{dx}\).

Substitute Back and Simplify

Substitute \(u = x^2 + y^2\) and \(\frac{du}{dx} = 2x + 2y \cdot \frac{dy}{dx}\) into the derivative of the logarithm:\[\frac{d}{dx}[\ln(x^2 + y^2)] = \frac{1}{x^2 + y^2}(2x + 2y \cdot \frac{dy}{dx})\]The right side's derivative is 0, so the equation we have now becomes:\[\frac{1}{x^2 + y^2}(2x + 2y \cdot \frac{dy}{dx}) = 0\]

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

We need to solve for \(\frac{dy}{dx}\). Start by multiplying both sides by \(x^2 + y^2\) to clear the fraction:\[2x + 2y \cdot \frac{dy}{dx} = 0\]Now, move \(2x\) to the other side:\[2y \cdot \frac{dy}{dx} = -2x\]Finally, divide by \(2y\) to solve for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{-2x}{2y} = \frac{-x}{y}\]

Key concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to find the derivative of composite functions. When you have a function that depends on another function, the chain rule helps you find the rate of change. In simpler terms, if you have a function within another function, you use the chain rule to differentiate it.

• Consider a composite function like \(f(g(x))\), where \(f\) is a function of \(g(x)\).
• To differentiate \(f(g(x))\), we use the formula \(f'(g(x)) \cdot g'(x)\).

When applying the chain rule in our exercise, we took \(u = x^2 + y^2\) and differentiated \(\ln(u)\) with respect to \(x\). By doing this, we used the chain rule to find the derivative of \(\ln(x^2 + y^2)\) where the inside function \(u\) is taken with its derivative \(\frac{du}{dx}\).
This rule is essential when dealing with complex functions because it allows us to break down the differentiation into manageable parts.

Derivative of Logarithmic Function

Differentiating a logarithmic function involves understanding its basic properties. The derivative of the natural logarithm function, such as \(\ln(x)\), is \(\frac{1}{x}\). When the argument of the logarithm is not just \(x\) but a more complex expression, such as \(x^2 + y^2\) in our exercise, we need to apply the chain rule as well.

• For \(\ln(u)\), the derivative with respect to \(x\) is \(\frac{1}{u} \cdot \frac{du}{dx}\).
• This means we must first find \(\frac{du}{dx}\) if \(u\) is a function of \(x\).

In our problem, we had the expression \(u = x^2 + y^2\). The derivative of \(\ln(x^2 + y^2)\) involved finding \(\frac{du}{dx}\), which required implicit differentiation for the \(y^2\) term and direct differentiation for the \(x^2\) term. This combination allowed us to correctly find the derivative of the logarithmic function used in the exercise.

Implicit Function Theorem

The implicit function theorem is an important concept that allows us to differentiate functions defined by equations, rather than explicit expressions. When a function is given implicitly, both \(x\) and \(y\) are typically dependent variables tied through an equation. Implicit differentiation uses this theorem to find derivatives.

• In implicit differentiation, we treat \(y\) as a function of \(x\).
• We then differentiate both sides of the equation with respect to \(x\).

In our exercise, we had the equation \(\ln(x^2 + y^2) = e\). The right side, \(e\), is a constant, so its derivative is 0. For the left side, we used implicit differentiation because \(y\) is not isolated. By finding \(\frac{du}{dx}\) with respect to both \(x\) and \(y\), we incorporated both direct and implicit derivatives, leading to the solution where \(\frac{dy}{dx} = \frac{-x}{y}\).
This methodology is crucial when working with equations where one variable is in terms of others, allowing us to uncover relationships between variables through their derivatives.