91Ó°ÊÓ

When faced with an expression like \( y \sin \left(\frac{1}{y}\right) \), we apply the product rule to differentiate it. The product rule is essential when you have two functions multiplied together. It states that if you have two functions, say \( u(x) \) and \( v(x) \), the derivative of their product is \( \frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v' \).
In our exercise, \( u = y \) and \( v = \sin \left(\frac{1}{y}\right) \). So, the derivative of \( y \sin \left(\frac{1}{y}\right) \) using the product rule is:

• Differentiate \( y \): \( \frac{d}{dx}(y) = \frac{dy}{dx} \).
• Keep \( \sin \left(\frac{1}{y}\right) \) unchanged: \( \sin \left(\frac{1}{y}\right) \).
• Differentiate \( \sin \left(\frac{1}{y}\right) \): We use the chain rule here (explained in the next section).
• Combine: \( \frac{dy}{dx} \cdot \sin \left(\frac{1}{y}\right) + y \cdot \frac{d}{dx}\left(\sin \left(\frac{1}{y}\right)\right) \).

This simplifies and structures the expression, making the differentiation process manageable.

The chain rule is a powerful tool in calculus used to differentiate composite functions. In simpler terms, if you have a function within another function, the chain rule helps you find the derivative. It says that if \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
In our equation, to differentiate \( \sin \left(\frac{1}{y}\right) \) using the chain rule, follow these steps:

• The outer function is \( \sin(u) \), where \( u = \frac{1}{y} \).
• The derivative of \( \sin(u) \) is \( \cos(u) \).
• The inner function is \( \frac{1}{y} \). Its derivative is \( -\frac{1}{y^2} \) times \( \frac{dy}{dx} \) due to the y-dependence.

Hence, the complete differentiation using the chain rule is:
\[ \frac{d}{dx}\left(\sin\left(\frac{1}{y}\right)\right) = \cos\left(\frac{1}{y}\right) \cdot \left(-\frac{1}{y^2}\right) \cdot \frac{dy}{dx} \]
Using the chain rule properly helps untangle complex expressions, making it simpler to move forward with implicit differentiation.

After applying the product and chain rules, you're often left with a differential equation. Solving these equations is crucial in calculus, especially when finding \( \frac{dy}{dx} \) in implicit contexts.
In our example, after differentiating both sides, we got the equation:
\[ \frac{dy}{dx} \cdot \left(\sin \left(\frac{1}{y}\right) - \frac{\cos \left(\frac{1}{y}\right)}{y} + x \right) = -y \]
To isolate \( \frac{dy}{dx} \):

• Collect terms involving \( \frac{dy}{dx} \) on one side of the equation.
• Factor out \( \frac{dy}{dx} \) from those terms.
• Rearrange equations to express \( \frac{dy}{dx} \) explicitly.

This way, you find \( \frac{dy}{dx} \) as a function of \( x \) and \( y \), which is
\[ \frac{dy}{dx} = \frac{-y}{\sin \left(\frac{1}{y}\right) - \frac{\cos \left(\frac{1}{y}\right)}{y} + x} \].
By doing so, the implicit relationship has transformed into an explicit derivative.