91Ó°ÊÓ

The chain rule is a fundamental concept in calculus, especially useful for finding derivatives of composite functions. When you have a function of a function, like in parametric equations, the chain rule helps simplify the differentiation process.

The general form of the chain rule states that if you have a function \( z = f(y) \), and \( y = g(x) \), then the derivative of \( z \) with respect to \( x \) is given by:

\[ \frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx} \]
For parametric equations like in our exercise, where \( x \) and \( y \) are both expressed in terms of another variable \( t \), the chain rule appears as:

\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]
This formula allows us to translate the derivatives from the parameter \( t \) to \( x \), making it easier to find how \( y \) changes with \( x \).

The quotient rule is crucial when dealing with derivatives of fractions, specifically where a quotient of two functions is involved. If \( y = \frac{u}{v} \) with both \( u \) and \( v \) as differentiable functions of \( t \), the derivative is given by:

\[ \frac{d}{dt} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dt} - u \cdot \frac{dv}{dt}}{v^2} \]
In this exercise, we applied the quotient rule to the derivative \( \frac{dy}{dx} = \frac{3t^2 + 4t}{3t^2 - 1} \) to find \( \frac{d(\frac{dy}{dx})}{dt} \).

The step-by-step method ensures the numerator changes by \( v \) and \( u \) appropriately, and divides them by the square of the original denominator ensuring a precise result. This rule is indispensable for calculating the second derivative, as it involves differentiating the already differentiated quotient.

Finding the first derivative, \( \frac{dy}{dx} \), of a parametric equation involves determining how \( y \) changes with \( x \). First, differentiate \( x \) and \( y \) with respect to \( t \):

- For \( x = t^3 - t \), we get \( \frac{dx}{dt} = 3t^2 - 1 \).
- For \( y = t^3 + 2t^2 \), we get \( \frac{dy}{dt} = 3t^2 + 4t \).

Then apply the chain rule:

\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2 + 4t}{3t^2 - 1} \]
This result shows the rate of change of \( y \) with respect to \( x \), expressing how the curve behaves at given points of \( t \).

The second derivative, \( \frac{d^2y}{dx^2} \), reflects the rate of change of the first derivative, showing how curvature changes. First, differentiate the first derivative \( \frac{dy}{dx} = \frac{3t^2 + 4t}{3t^2 - 1} \) with respect to \( t \) using the quotient rule.

Here's how:

- Set \( u = 3t^2 + 4t \) and \( v = 3t^2 - 1 \).
- Calculate \( \frac{du}{dt} = 6t + 4 \) and \( \frac{dv}{dt} = 6t \).

Apply the quotient rule:

\[ \frac{d(\frac{dy}{dx})}{dt} = \frac{(3t^2 - 1)(6t + 4) - (3t^2 + 4t)(6t)}{(3t^2 - 1)^2} \]
Finally, use the chain rule again to find \( \frac{d^2y}{dx^2} \):

\[ \frac{d^2y}{dx^2} = \frac{\frac{d(\frac{dy}{dx})}{dt}}{\frac{dx}{dt}} = \frac{(3t^2 - 1)(6t + 4) - (3t^2 + 4t)(6t)}{(3t^2 - 1)^3} \]
This captures not only the rate of change but the concavity of the curve at different \( t \) values.