91Ó°ÊÓ

The chain rule is a fundamental concept in calculus used when differentiating a function composed of multiple smaller functions. Imagine it like peeling layers of an onion; you need to take each layer into account while differentiating.
\(e^{xy}\) is a composition of two simple functions, the exponential function \(e^u\) and the product \(xy\). It requires the chain rule because of this nested structure. Here's how it works in our exercise:

• Identify the inner and outer functions: Inner is the multiplicative \(xy\) and outer is the exponential function \(e^u\).


• Apply the chain rule: First differentiate the outer function \(e^u\) with respect to \(u\), giving \(e^u\), and then multiply by the derivative of the inner function \(xy\) with respect to \(x\).

In doing so, the derivative converts into an expression involving both x and y: \[ \frac{d}{dx}(e^{xy}) = e^{xy} \cdot (y + x \cdot \frac{dy}{dx}) \] where \(y + x \cdot \frac{dy}{dx}\) comes from differentiating \(xy\). The chain rule helps to manage these layers, giving a complete derivative function for composite structures.

In implicit differentiation, we differentiate one variable in terms of another, often \(x\), even when \(y\) is not isolated. This approach becomes essential when equations aren’t explicitly solved for y. Here’s a simple breakdown:
To differentiate explictly, every term must be regarded in terms of x. In our problem:

• We follow the rules for each type of term: For the left side involving \(e^{xy}\), apply the chain rule.


• For the right side \(y^2\), regard \(y\) as a function of \(x\). By doing so, we differentiate as \(2y \cdot \frac{dy}{dx}\) instead of treating \(y\) as a constant.


• The term \(-x\) becomes \(-1\), as its derivative concerning \(x\) is a constant response to a linear variable.

Implicit differentiation allows us to uncover how each variable contributes to changes in the function as a whole.

Factoring, in the context of solving derivative equations, involves rearranging and simplifying expressions to isolate certain terms. It becomes a crucial step when aiming to solve for derivatives, such as \( \frac{dy}{dx} \).
In our exercise, once the equations are differentiated, there's a mix of like terms involving \( \frac{dy}{dx} \). Factoring these terms involves:

• Rewriting the equation in a form where \( \frac{dy}{dx} \) can be factored out. This is achieved by grouping so the expressions are combined and separated from non-\( \frac{dy}{dx} \) terms.


• Isolating \( \frac{dy}{dx} \) makes solving for it easier, especially in complex implicit differentiation scenarios. The equation turns into:

\[ \left( e^{xy} \cdot x - 2y \right) \cdot \frac{dy}{dx} = -1 - e^{xy} \cdot y \] From here, the solution for \( \frac{dy}{dx} \) can be achieved by dividing by the expression \((e^{xy} \cdot x - 2y)\), leading to our result:
\[ \frac{dy}{dx} = \frac{-1 - e^{xy} \cdot y}{e^{xy} \cdot x - 2y} \] This process simplifies complex derivatives into manageable forms, highlighting the usefulness of factoring.