91Ó°ÊÓ

The chain rule is a vital part of calculus, allowing us to differentiate functions that are composed of other functions. When you have a function nested inside another, you use the chain rule to take the derivative.
Simply, the chain rule is stated as:

• If you have a composite function \(f(g(x))\), the derivative is calculated as \( f'(g(x)) \cdot g'(x) \).

In the context of implicit differentiation, where we have an equation like \( y^2 \), applying the chain rule helps us find the derivative of \( y^2 \) as it changes with \( x \). Since \( y \) itself is a function of \( x \), we treat \( y \) as \( u \), making the derivative \( 2y \cdot \frac{dy}{dx} \).
This method is crucial because many equations that involve both \( x \) and \( y \) simultaneously aren't explicitly given in terms of \( y \). Thus, using the chain rule allows us to differentiate such equations implicitly.

The product rule is your go-to tool for finding derivatives when dealing with products of two functions. If you have two differentiable functions, say \( u(x) \) and \( v(x) \), their product \( u(x)v(x) \) is differentiated using:

• The product rule formula: \(( uv )' = u'v + uv'\).

In implicit differentiation, you often find functions intertwined, like \( x \ln y \). For this, the product rule helps break it down:

• Differentiate \( x \, (\text{considered as} \, u) \), which gives \( 1 \).
• Differentiate \( \ln y \, (\text{considered as} \, v) \), which needs the chain rule: \( \frac{1}{y} \cdot \frac{dy}{dx} \).

Thus, for \( x \ln y \), the derivative becomes \( \ln y \cdot 1 + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} \).
Applying the product rule in this implicit context allows us to respect the intricacies of how functions interplay within an equation.

Once you have differentiated both sides of an equation using implicit differentiation, you arrive at a point where you need to solve for \( \frac{dy}{dx} \). This is about isolating \( \frac{dy}{dx} \) on one side of the equation.

• Reorganize the equation by combining the terms involving \( \frac{dy}{dx} \).
• In the current example, \( 2y - \frac{x}{y} \) is a common factor of such terms.
• Factor out \( \frac{dy}{dx} \) as shown: \( \left( 2y - \frac{x}{y} \right) \cdot \frac{dy}{dx} = \ln y \).

Finally, divide both sides of the equation by the factor which multiplies \( \frac{dy}{dx} \) to solve for it:

• \( \frac{dy}{dx} = \frac{\ln y}{2y - \frac{x}{y}} \).

This process provides the rate of change of \( y \) with respect to \( x \). Being able to manipulate and rearrange terms algebraically is essential in finding the implicit derivative.