91Ó°ÊÓ

The chain rule is a powerful tool in calculus that helps us differentiate composite functions. In simple terms, it states that if you have a function inside another function, you need to differentiate both the inner and the outer functions.
For instance, when differentiating \( \sqrt{y} \) with respect to \( x \), we consider \( y \) as a function of \( x \), say \( y = f(x) \). The chain rule then tells us to first take the derivative of \( \sqrt{y} \) with respect to \( y \), and then multiply by the derivative of \( y \) with respect to \( x \).
Mathematically:
\[ \frac{d}{dx} \sqrt{y} = \frac{d}{dy} \sqrt{y} \cdot \frac{dy}{dx} = \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} \] For \( \sqrt{x} \), since \( x \) is directly your variable, it's more straightforward: \[ \frac{d}{dx} \sqrt{x} = \frac{1}{2\sqrt{x}} \]. In the exercise, you may have noted that differentiating the left side \( \sqrt{x} + \sqrt{y} \) used the chain rule only for \( \sqrt{y} \), while \( \sqrt{x} \) was directly differentiated without it.

Derivatives represent the rate of change of a function. In practice, they tell you how a function changes as its inputs change.
For the function \( \sqrt{x} \), differentiating it gives you a sense of how \( \sqrt{x} \) changes as \( x \) changes.
Here’s a quick rundown of the steps to differentiate \( \sqrt{x} \):

• Recognize \( \sqrt{x} \) as \( x^{1/2} \).
• Apply the power rule to get \( \frac{1}{2} x^{-1/2} \) or \( \frac{1}{2\sqrt{x}} \).

Differentiating implicit functions, like in \( \sqrt{y} \), means considering \( y \) as a dependent function of \( x \).
The chain rule comes into play here, and the steps involve multiple parts:

• For \( \sqrt{y} \), take the derivative with respect to \( y \), yielding \( \frac{1}{2\sqrt{y}} \).
• Then multiply by \( \frac{dy}{dx} \), because \( y \) itself is a function of \( x \).

This keeps the derivative consistent within the context of the entire expression.

After differentiating implicitly, you'll often need to isolate \( \frac{dy}{dx} \), which involves algebraic manipulation.
Taking the combined differentiated equation:
\[ \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0 \]
Here are the steps to solve for \( \frac{dy}{dx} \):

• First, subtract \( \frac{1}{2\sqrt{x}} \) from both sides: \[ \frac{1}{2\sqrt{y}} \frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \]
• Then, multiply both sides by \( 2\sqrt{y} \): \[ \frac{dy}{dx} = -2\sqrt{y} \cdot \frac{1}{2\sqrt{x}} \]
• Finally, simplify to get: \[ \frac{dy}{dx} = -\sqrt{y} \cdot \frac{1}{\sqrt{x}} = -\sqrt{\frac{y}{x}} \]

These steps ensure that you've properly isolated \( \frac{dy}{dx} \) and expressed it in terms of known quantities.