91Ó°ÊÓ

Implicit differentiation is a technique used when it is difficult or impossible to solve an equation for one variable in terms of another. When you have a function defined by an equation like \( 2xy + y = 1 \), you can't easily express \( y \) in terms of \( x \).
Instead of solving for \( y \) initially, you differentiate both sides of the equation directly with respect to \( x \).
This method involves treating \( y \) as a function of \( x \) (even if it is not explicitly given that way), and using the chain rule when differentiating terms involving \( y \).

• Differentiate each side with respect to \( x \).
• Apply the chain rule to terms involving \( y \), where \( \frac{dy}{dx} \) is considered the derivative of \( y \) with respect to \( x \).
• Use algebraic manipulation to solve for \( \frac{dy}{dx} \).

This approach allows you to determine how \( y \) changes in response to changes in \( x \) without explicitly solving the function \( y = f(x) \).

When discussing the instantaneous rate of change, we are referring to how a quantity changes at an exact point. For functions, this is essentially what a derivative represents—the slope of the tangent line at a point on the curve.
For example, when you calculate \( \frac{dy}{dx} \) for \( x = 0.5 \), you are finding how rapidly \( y \) changes as \( x \) changes, precisely at \( x = 0.5 \).
This can be visualized as a snapshot of the function's behavior:

• The derivative \( \frac{dy}{dx} \) gives you the slope of the tangent line at that point.
• The steeper the slope, the faster the rate of change.

In real-life applications, this could represent anything from speed (change of position with respect to time) to how quickly a population is growing or shrinking at a specific moment.

The product rule in calculus is essential when you differentiate products of two or more functions of \( x \).
In our example of \( 2xy + y = 1 \), the product of \( 2x \) and the function \( y \) requires special attention.
To find the derivative of \( 2xy \), you cannot simply differentiate \( 2x \) and \( y \) independently.
Instead, apply the product rule which states:

• If \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative of their product \( u(x) v(x) \) is given by:
  \[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \]

In this case:

• \( u(x) = 2x \), and \( u'(x) = 2 \)
• \( v(x) = y \), and \( v'(x) = \frac{dy}{dx} \)

The product rule results in \( 2 \cdot y + 2x \cdot \frac{dy}{dx} \), allowing us to properly account for both functions' rates of change in the derivative.

The first derivative, denoted as \( \frac{dy}{dx} \), tells us about the rate at which \( y \) changes with respect to \( x \).
This piece of information is crucial in understanding the behavior of functions. It informs us whether a function is increasing or decreasing at any given point.

• A positive first derivative means the function is increasing at that point.
• A negative derivative indicates a decreasing function at that specific \( x \)-value.
• If the first derivative equals zero, the function might be at a local maximum or minimum, depending on further analysis.

In our exercise, we calculated \( \frac{dy}{dx} = \frac{-1}{2} \) when \( x = 0.5 \).
This means the function \( y \) is decreasing at this point because the slope of the tangent line is negative, indicating a downward direction.