91Ó°ÊÓ

Leibnizian notation is a very convenient way to perform implicit differentiation. It uses the notation \(\frac{dy}{dx}\) to represent the derivative of \(y\) with respect to \(x\). This notation clearly shows which variable is being differentiated with respect to the other.
In the given problem, we start with the equation: \[ y^4 + y^3 + xy = 1 \] We then differentiate both sides with respect to \(x\) using Leibnizian notation. This notation helps us manage the calculation step-by-step and keeps track of which variables we are differentiating.

The chain rule is a vital tool in differentiation, especially when dealing with implicit differentiation. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In symbols, if \(u=f(v(x))\), then \(\frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx}\).
In our example, the terms \(y^4\) and \(y^3\) need the chain rule because \(y\) is a function of \(x\). So we write these terms as follows:
- For \(y^4\): \[ \frac{d}{dx}(y^4) = 4y^3 \frac{dy}{dx} \]
- For \(y^3\): \[ \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx} \]
This rule allows us to correctly include the derivative of \(y\) with respect to \(x\) in our calculations.

The product rule is essential when differentiating expressions where two functions are multiplied together. According to the product rule, if we have two functions \(u(x)\) and \(v(x)\), their product's derivative is given by \(\frac{d}{dx}[u(x) \cdot v(x)] = u'(x)v(x) + u(x)v'(x)\).
In our problem, the term \(xy\) requires the product rule because it involves the product of \(x\) and \(y\) (which is a function of \(x\)). Therefore:
\( \frac{d}{dx}(xy) = x \frac{dy}{dx} + y \)
This helps us accurately differentiate each part of the equation and capture the contributions of both \(x\) and \(y\).

Differentiation is the process of finding the rate at which a function is changing at any given point. In the context of implicit differentiation, it involves differentiating equations where the dependent variable is not isolated.
Starting with our example equation: \[ y^4 + y^3 + xy = 1 \],
we differentiate each term separately with respect to \(x\). For terms involving \(y\), we apply the chain rule, and for products of \(x\) and \(y\), we use the product rule. After differentiating, we combine the terms: \[ 4y^3 \frac{dy}{dx} + 3y^2 \frac{dy}{dx} + y + x \frac{dy}{dx} = 0 \]
Finally, we solve for \( \frac{dy}{dx} \) by isolating it on one side of the equation: \[ \frac{dy}{dx} = \frac{-y}{4y^3 + 3y^2 + x} \]. This gives us the rate of change of \(y\) with respect to \(x\).