91Ó°ÊÓ

When differentiating products of functions, the product rule is an invaluable tool. The rule states that the derivative of a product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second. In shorthand, if you have two functions, say \( u \) and \( v \), then the derivative \( \frac{d}{dx}(uv) \) is given by:

• \( u'v + uv' \)

In the original exercise, while finding the derivative of \( x(y - 1)^2 \), see \( x \) as \( u \) and \( (y - 1)^2 \) as \( v \). Then use the product rule to separate each term to make the implicit differentiation easier.
First, differentiate \( x \) to get \( u' = 1 \). Then, for \( (y-1)^2 \), the derivative is not as straightforward, so the chain rule is next needed.

The chain rule is crucial when taking derivatives of functions that are compositions of other functions. It allows us to differentiate the inside and outside of the function separately where necessary.
In simple terms, if you have a function \( g(f(x)) \), the chain rule lets us express the derivative as \( g'(f(x)) \cdot f'(x) \).
In the exercise, the function \( (y-1)^2 \) requires the chain rule. Treat \( y-1 \) as an inner function, \( u(y) \), so its derivative requires:

• First, differentiate the outside: \( 2(y-1) \).
• Then, multiply by the derivative of the inside: \( \frac{dy}{dx} \).

This results in the derivative \( v' = 2(y-1) \cdot \frac{dy}{dx} \). This allows you to manage and solve for \( \frac{dy}{dx} \) for equations using implicit differentiation.

Derivatives are a fundamental part of calculus, addressing how a function changes as its inputs change. Implicit differentiation is especially useful when dealing with equations where \( y \) is not isolated.
To solve such an equation to find \( \frac{dy}{dx} \), our previous steps combined as:

• First, apply the product and chain rules to get \((y-1)^2 + x \cdot 2(y-1) \cdot \frac{dy}{dx} = 0 \).
• Second, solve for \( \frac{dy}{dx} \) by isolating it on one side of the equation.

To isolate \( \frac{dy}{dx} \), move terms involving just \( y \) to one side and the terms with \( \frac{dy}{dx} \) to the other. For the exercise, this led to:\[ \frac{dy}{dx} = \frac{-(y-1)}{2x} \]Simplifying these terms often involves factoring or cancelling out like terms, just like dividing \((y-1)\) from both the numerator and the denominator. This solves the implicit equation for the derivative with respect to \( x \).