91Ó°ÊÓ

When working with implicit differentiation, the chain rule becomes a crucial tool. This rule is used when differentiating composite functions. Here, we treat one variable as a function of another. In our exercise, this means treating \( y \) as a function of \( x \).
To understand how the chain rule works, consider a composite function \( f(g(x)) \). The chain rule suggests that to differentiate \( f(g(x)) \), you first differentiate \( f \) concerning \( g \) (i.e., \( f'(g(x)) \)), and then multiply by the derivative of \( g \) concerning \( x \) (i.e., \( g'(x) \)):

• \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \).

Applying this in our exercise, for a term like \( y^2 \), we first differentiate with respect to \( y \), giving \( 2y \), and then multiply by \( \frac{dy}{dx} \) because we need the derivative with respect to \( x \). Hence, the full differentiation of \( y^2 \) is \( 2y \cdot \frac{dy}{dx} \). It's essential to always follow this sequence to apply the chain rule correctly.

The product rule is another fundamental concept in differentiation, particularly useful when dealing with products of two functions. For a product of two functions \( u(x) \) and \( v(x) \), the product rule states:

• \( \frac{d}{dx}[uv] = u'v + uv' \)

In our exercise, the term \( 2xy^2 \) involves a product of \( 2x \) and \( y^2 \). The product rule requires us to differentiate both parts and sum the results.
First, differentiate \( 2x \), which gives \( 2 \), keeping \( y^2 \) as it is. Then, differentiate \( y^2 \) using the chain rule, as described earlier, resulting in \( 2y \cdot \frac{dy}{dx} \). Now, sum these products:

• \( 2 \cdot y^2 + 2x \cdot 2y \cdot \frac{dy}{dx} = 2y^2 + 4xy \cdot \frac{dy}{dx} \).

This method allows us to fully differentiate terms that involve multiplication of functions, a frequent occurrence in implicit differentiation.

Differentiation is the process of finding the derivative of a function, which represents the rate at which the function changes. In implicit differentiation, this process involves treating one variable as a function of another and differentiating all terms with respect to a single variable, like \( x \).
This is particularly useful when dealing with equations not easily solved for one variable in terms of another. In our exercise, the whole equation \( x^3 + 2xy^2 + y^3 = 1 \) needs to be differentiated concerning \( x \).

• For \( x^3 \), we use simple differentiation to get \( 3x^2 \).
• For \( 2xy^2 \), we apply both the product rule and chain rule, already discussed.
• For \( y^3 \), the chain rule gives \( 3y^2 \cdot \frac{dy}{dx} \).

By combining all the derivatives, we simplify the expression, eventually solving for \( \frac{dy}{dx} \), which is the overall objective in implicit differentiation. Understanding these steps in differentiation ensures you can handle more complex differentiation problems effectively.