91Ó°ÊÓ

When dealing with a problem that requires implicit differentiation, finding the first derivative is your starting point. Implicit differentiation is a useful technique, particularly when dealing with equations where the variable you want to differentiate with respect to (often "x") and the other variable (often "y") are intertwined.In the provided exercise, we are tasked with differentiating the equation: \[ y^2 + 2y - 2x = 1 \]with respect to "x". To do this, differentiate each term:

• For \( y^2 \), apply the chain rule to get: \( 2y \frac{dy}{dx} \).
• For \( 2y \), it becomes: \( 2 \frac{dy}{dx} \).
• For \(-2x\), the derivative is simply: \(-2\).

By gathering these differentiated terms, we derive the equation:\[ 2y \frac{dy}{dx} + 2 \frac{dy}{dx} - 2 = 0 \]Now, solve for \( \frac{dy}{dx} \) by factoring it out and isolating:\[ (2y + 2) \frac{dy}{dx} = 2 \]Then divide by \(2y + 2\):\[ \frac{dy}{dx} = \frac{2}{2(y+1)} = \frac{1}{y+1} \]This expression gives us the first derivative, which delivers rates of change for the variable "y" in terms of "x".

Once we have found the first derivative using implicit differentiation, the next step is to determine the second derivative, \( \frac{d^2y}{dx^2} \). The second derivative provides deeper insight into the concavity and curvature of a curve, or how the rate of change of a function itself changes over time.Starting with the first derivative \( \frac{dy}{dx} = \frac{1}{y+1} \), differentiate with respect to "x" to find the second derivative. This involves using the chain rule:\[ \frac{d}{dx}\left( \frac{1}{y+1} \right) = -\frac{1}{(y+1)^2} \cdot \frac{dy}{dx} \]Substitute \( \frac{dy}{dx} = \frac{1}{y+1} \) back into the equation:\[ \frac{d^2y}{dx^2} = -\frac{1}{(y+1)^2} \cdot \frac{1}{y+1} = -\frac{1}{(y+1)^3} \]This second derivative tells us how the slope (first derivative) changes, essentially indicating the bend or change in curvature of the function.

The chain rule is a central concept when working with implicit differentiation, particularly useful when differentiable functions are composed in a nested way. In simple terms, the chain rule helps us differentiate composite functions: functions of one function.In our exercise, applying the chain rule is crucial for finding both derivatives. When we differentiate terms like \( y^2 \) and \( 2y \) with respect to "x", we apply the chain rule:

• For \( y^2 \), it becomes \( 2y \frac{dy}{dx} \). The outer function is \( y^2 \), and the inner function is "y" as a function of "x".
• Similarly, for \( 2y \), differentiation results in \( 2 \frac{dy}{dx} \).

For the second derivative, the chain rule is applied once more:When differentiating \( \frac{dy}{dx} = \frac{1}{y+1} \), we must multiply the derivative of the outer function \( -\frac{1}{(y+1)^2} \) by the derivative of the inner function \( \frac{dy}{dx} \), producing:\[ -\frac{1}{(y+1)^3} \]The chain rule thus serves as an indispensable tool for managing layers of functions within one another, aiding in finding accurate derivatives.