91Ó°ÊÓ

Implicit differentiation is a technique used in calculus when a function is not explicitly solved for one variable in terms of others. In some cases, you have equations where one of the variables, often denoted as \( y \), is a function of another variable, such as \( x \), and they are mixed together in an equation.
For example, in the equation \( y^2 - x^3 = 1 \), \( y \) is not isolated, meaning it's intertwined with \( x \).
When differentiating implicitly, we apply the derivative to both sides of the equation with respect to \( x \), considering \( y \) as a function of \( x \).

• The power rule can be used, for instance \( \frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx} \), where \( \frac{dy}{dx} \) represents the derivative of \( y \) with respect to \( x \).
• Similarly, you differentiate \( -x^3 \) using normal rules, resulting in \( -3x^2 \).

This "implicit" approach allows us to find the derivative even when the variables are interdependent in complex equations.

In calculus, the derivative is a core concept that measures how a function changes as its input changes. When you find the derivative of a function, you're essentially finding the "rate of change" or the "slope" of the function at a given point.

In the context of the exercise, after differentiating the equation \( y^2 - x^3 = 1 \) implicitly, we get a new equation: \( 2y \frac{dy}{dx} - 3x^2 = 0 \).
In this equation, \( \frac{dy}{dx} \) represents the derivative of \( y \) with respect to \( x \). This is what we're solving for:

• The term \( 2y \cdot \frac{dy}{dx} \) comes from differentiating \( y^2 \) with the chain rule.
• \( -3x^2 \) is straightforward as it's the derivative of \( -x^3 \).

Solving for \( \frac{dy}{dx} \) gives insight into how \( y \) changes as \( x \) changes.

Solving equations is a fundamental task when it comes to calculus and derivatives. Once we have differentiated and obtained our derivative equation, it is crucial to solve for \( \frac{dy}{dx} \).

The equation \( 2y \frac{dy}{dx} = 3x^2 \) must have \( \frac{dy}{dx} \) isolated. To do this, rearrange the equation and solve for the derivative:

• Add \( 3x^2 \) to both sides first to separate terms involving \( \frac{dy}{dx} \).
• Then divide by \( 2y \) to make \( \frac{dy}{dx} \) stand alone: \( \frac{dy}{dx} = \frac{3x^2}{2y} \).

Once \( \frac{dy}{dx} \) is expressed in terms of \( x \) and \( y \), substitute any given values of \( x \) and \( y \), as shown in the example with \( x=2 \) and \( y=3 \), enabling finding the specific value of the derivative at those points. This step solidifies understanding of how substitution works in finding specific slope or rate of change in functions.