91Ó°ÊÓ

When you find the second derivative, you are essentially looking for the rate at which the first derivative changes. In implicit differentiation, this becomes crucial, especially when dealing with functions defined implicitly rather than explicitly. Here, we started with the equation \(x^4 + y^4 = a^4\). To find the second derivative \(y^{\prime\prime}\), we must first differentiate the equation with respect to \(x\) to obtain the first derivative \(\frac{dy}{dx}\). This step uses the chain rule since \(y\) is a function of \(x\), even though it is not specified explicitly. Once we have \( \frac{dy}{dx} = -\frac{x^3}{y^3} \), we differentiate again. The process involves applying the quotient rule, since our expression is a fraction. In the solution, after differentiating again, we've arrived at the second derivative: \[ y'' = \frac{-3x^2 y^6 - 3x^6}{y^9} \].The second derivative gives us insights into the concavity of the function and how the slope of the tangent line changes at each point.

The chain rule is a fundamental tool in calculus used for differentiating composite functions. In our exercise, we encounter it when differentiating \(y^4\) with respect to \(x\). Because \(y\) is actually a function of \(x\), differentiating \(y^4\) requires us to apply the chain rule. In broader terms, if you have a function \(f(g(x))\), the chain rule states that the derivative \(f'(g(x)) \cdot g'(x)\) gives us the rate of change of \(f\) with respect to \(g\) times the rate of change of \(g\) with respect to \(x\).

• For our problem: differentiate \(y^4\), resulting in \(4y^3 \frac{dy}{dx}\).
• This calculation reflects the notion that differentiating each part respects the dependency of \(y\) on \(x\).
• Makes implicit differentiation possible by considering every appearance of \(y\) as \(y(x)\).

Understanding the chain rule is essential for implicit differentiation, allowing us to handle expressions where variables are interdependent.

The quotient rule is used to differentiate functions that are divided by one another. In our exercise, we used the quotient rule to differentiate \(\frac{dy}{dx} = -\frac{x^3}{y^3}\) in order to find \(y^{\prime\prime}\). The quotient rule states that for two functions \(u(x)\) and \(v(x)\), the derivative of their quotient \(\frac{u}{v}\) is given by:\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2} \] In our exercise:

• Set \(u = -x^3\) and \(v = y^3\).
• The derivatives: \(u' = -3x^2\) and \(v' = 3y^2 \frac{dy}{dx}\).
• Apply the quotient rule to find \(y'' = \frac{y^3(-3x^2) - (-x^3)(3y^2 \frac{dy}{dx})}{y^6}\). This expression shows how each part of the fraction changes and impacts the derivative.

The quotient rule is indispensable for complex fractions and ensures accurate differentiation when both numerator and denominator are variable-functions.