91Ó°ÊÓ

In calculus, the chain rule is an essential concept, especially when dealing with implicit differentiation. It allows us to find the derivative of compositions of functions. Simply put, if you have a function inside another function, the chain rule helps you differentiate the outer function and multiply it by the derivative of the inner function. This is incredibly useful when the variable you're differentiating with respect to is not the explicit variable in the function.

For instance, in our original problem, we use the chain rule to differentiate \(\frac{1}{y}\) with respect to \(x\). Although \(y\) is not directly a function of \(x\), we treat it as one by applying the chain rule. We end up with \(-\frac{1}{y^2} \cdot \frac{dy}{dx}\) because we differentiate \(\frac{1}{y}\) as \(y^{-1}\) and then multiply by \(\frac{dy}{dx}\), acknowledging that \(y\) itself depends on \(x\).

• Differentiating the outer function: treat \(y^{-1}\) as a typical power of a variable.
• Multiplying by the derivative of the inner function: \(\frac{dy}{dx}\), since \(y\) depends on \(x\).

This method simplifies finding derivatives when variables are interdependent.

The derivative is a fundamental tool in calculus that measures how a function changes as its input changes. Essentially, it's the "rate of change" or "slope" of a function at a particular point. For a given functional expression, the derivative can be computed directly if the relationship between variables is explicit, or indirectly through implicit differentiation when they are not.

In explicit differentiation, you directly apply derivative rules like the power rule or quotient rule. However, in implicit differentiation, as seen in the original exercise, we differentiate both sides of an equation that involves two related variables. Here's how it applies:

• Start by differentiating each term with respect to \(x\), even if one or more terms are expressed in terms of \(y\).
• Use derivative rules to simplify the derivatives of simple fractions, such as \(\frac{1}{x}\) and \(\frac{1}{y}\).

By knowing how to take derivatives, especially with techniques like implicit differentiation, you can handle equations that don’t easily separate into functions of a single variable.

Once derivatives are calculated, equation solving comes into play. Solving equations is about isolating a variable to find its expression in terms of others. After using implicit differentiation, the next logical step is to solve for \( \frac{dy}{dx} \). This involves algebraic manipulation.

In the given exercise, after obtaining the equation from differentiating both sides,\[ -\frac{1}{y^2} \cdot \frac{dy}{dx} - \frac{1}{x^2} = 0, \]you need to solve for \( \frac{dy}{dx} \). Here's how it unfolds:

• Move terms to isolate the term containing \( \frac{dy}{dx} \), by adding or subtracting terms.
• Each step involves clear and simple operations, such as multiplying both sides by \(-y^2\) to eliminate fractions.
• Finally, rearrange to find \( \frac{dy}{dx} = \frac{y^2}{x^2} \).

These steps ensure a systematic way of finding the relationship between the rates of change of related variables. Once you understand the process of solving equations after differentiation, tackling implicit differentiation problems becomes much easier.