91Ó°ÊÓ

The second derivative provides information about how the rate of change is itself changing. If the first derivative \( \frac{dy}{dx} \) represents the slope or gradient, then the second derivative \( \frac{d^2y}{dx^2} \) tells us about the curvature or concavity of the function.
When we computed the second derivative in the exercise, we applied implicit differentiation to both sides of the original equation. This helps us explore not just how \(y\) changes with \(x\), but how swiftly those changes accelerate or decelerate.
In our problem, we started with the equation \( 2x^2 + y^2 = 4 \) and used implicit differentiation to find \( \frac{dy}{dx} = -\frac{2x}{y} \). The next step was to find \( \frac{d^2y}{dx^2} \), the rate at which the slope \( \frac{dy}{dx} \) changes, by differentiating \( \frac{dy}{dx} \) again with respect to \(x\). By applying the quotient rule, we arrived at the expression \( \frac{d^2y}{dx^2} = \frac{2x}{y^2} \). This expression shows how the rate of slope change is influenced by both \(x\) and \(y\).

The quotient rule is a vital tool in calculus for finding the derivative of a fraction where both the numerator and the denominator are functions of \(x\).
When you have an equation like \( \frac{u(x)}{v(x)} \), where both \(u\) and \(v\) are functions of \(x\), the derivative \( \frac{d}{dx}\left(\frac{u}{v}\right) \) is given by:
\[ \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2} \]
In our exercise, we needed to differentiate \( -\frac{2x}{y} \) to find the second derivative. Here, \(u = -2x\) and \(v = y\). By applying the quotient rule, we substituted these into the formula to get \[ \frac{y(-2) - (-2x) \cdot \frac{dy}{dx}}{y^2} \]. By simplifying further, and using the known value of \( \frac{dy}{dx}\), we arrived at the result of our second derivative.
The quotient rule ensures that we properly account for how changes in both numerator and denominator affect the rate of change in the whole fraction.

The power rule is a basic differentiation rule used when finding the derivative of a function with a power. If you have a term \(x^n\), the derivative is \(nx^{n-1}\). This rule simplifies the differentiation process greatly.
In our original exercise, one of the first steps involved differentiating \(2x^2\). Applying the power rule here, we took the exponent 2, multiplied it by the coefficient 2, resulting in \(4x\).
Similarly, for \(y^2\), the derivative with respect to \(x\) was \(2y\cdot \frac{dy}{dx}\). Notice how applying the chain rule here included \(\frac{dy}{dx}\) as \(y\) is dependent on \(x\).
By leveraging the power rule, we quickly turn complex-looking equations into simpler derivatives, helping us embark on the journey to higher derivatives like \( \frac{d^2y}{dx^2} \). Utilizing the power rule lays the groundwork and provides clarity in the differentiation process, making it a pivotal tool in calculus.