91Ó°ÊÓ

Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable. In this problem, we have the equation \(2x^2 + y^2 = 4\), which involves both \(x\) and \(y\). To begin with implicit differentiation, we take the derivative of each term with respect to \(x\) on both sides of the equation.

• Remember that \(y\) is a function of \(x\), so differentiating \(y^2\) with respect to \(x\) will utilize the chain rule.
• Differentiating \(2x^2\), we get \(4x\).
• For \(y^2\), applying the chain rule gives \(2yy'\).

The right side derivative of the constant 4 is 0. Hence, the equation becomes:\[ 4x + 2yy' = 0 \]We can solve for \(y'\) and get \(-\frac{2x}{y}\), which represents the rate of change of \(y\) with respect to \(x\). Implicit differentiation helps us find \(y'\) without needing to solve the original equation for \(y\).

The chain rule is a fundamental principle in calculus for differentiating compositions of functions. It is especially useful when dealing with nested functions. Suppose, for example, \(z = f(g(x))\), the derivative \(\frac{dz}{dx}\) is the product of the outer derivative evaluated at the inner function and the derivative of the inner function.In our example, \(2x^2 + y^2 = 4\), the chain rule is applied when differentiating \(y^2\). Assuming \(y\) is a function of \(x\), when we differentiate \(y^2\), we write:

• First, take the derivative of the outside, \(2y\), treating \(y\) as variable.
• Then, multiply by \(y'\) since \(y\) is a function of \(x\).

This results in the derivative \(2yy'\). Essentially, the chain rule allows us to account for how changes in \(x\) affect \(y\), and subsequently, \(y^2\). It is invaluable in situations where a direct derivative isn't straightforward due to nested or linked functions.

The quotient rule is necessary for differentiating ratios of functions. If you have a function presented as a fraction, \(\frac{u}{v}\), the derivative can be found using the formula:\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2} \]In our problem, after determining the first derivative \(y' = -\frac{2x}{y}\), we use the quotient rule to differentiate this to find \(y''\). In this expression:

• Identify \(u = -2x\) and \(v = y\).
• Compute \(u'\), the derivative of \(-2x\), which is \(-2\).
• Compute \(v'\), which is \(y'\), knowing that \(v = y\).

Plug these into the quotient rule gives:\[ y'' = \frac{y(-2) - (-2x)y'}{y^2} \]Substituting the expression for \(y'\) and simplifying results in:\[ y'' = \frac{-2y + 4x^2}{y^3} \]The quotient rule helps find the second derivative, capturing the rate of change of the change, which is essential for understanding curves involving quota-based relationships.