Implicit differentiation is a technique used to find the derivative of an equation when it's not explicitly solved for one variable in terms of another. In other words, implicit differentiation is applied when both variables (typically \(x\) and \(y\)) are mixed together in an equation rather than having \(y\) isolated as a function of \(x\).

### Steps to Implicitly Differentiate:

1. **Differentiate both sides of the equation with respect to \(x\)**: You treat \(y\) as a function of \(x\) (i.e., \(y = f(x)\)), so when differentiating terms involving \(y\), you apply the chain rule.

   - For example, the derivative of \(y^2\) with respect to \(x\) is:
     \[
     \frac{d}{dx}(y^2) = 2y \frac{dy}{dx}
     \]
     This happens because you're treating \(y\) as a function of \(x\), so the chain rule applies.

2. **Solve for \(\frac{dy}{dx}\)**: After differentiating both sides of the equation, you'll have an expression involving \(\frac{dy}{dx}\). The goal is to isolate \(\frac{dy}{dx}\), which represents the derivative of \(y\) with respect to \(x\).

### Example:
Consider the equation of a circle: 
\[
x^2 + y^2 = 25
\]

1. Differentiate both sides with respect to \(x\):
   \[
   \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)
   \]
   - \(\frac{d}{dx}(x^2) = 2x\)
   - \(\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}\) (using the chain rule)
   - \(\frac{d}{dx}(25) = 0\) (since 25 is a constant)

   So the equation becomes:
   \[
   2x + 2y \frac{dy}{dx} = 0
   \]

2. Solve for \(\frac{dy}{dx}\):
   \[
   2y \frac{dy}{dx} = -2x
   \]
   \[
   \frac{dy}{dx} = \frac{-2x}{2y} = \frac{-x}{y}
   \]

Thus, the derivative of \(y\) with respect to \(x\) for the equation \(x^2 + y^2 = 25\) is \(\frac{dy}{dx} = \frac{-x}{y}\).

### Key Points:
- Use implicit differentiation when \(y\) cannot be easily isolated in the equation.
- Apply the chain rule when differentiating terms with \(y\).
- After differentiation, solve for \(\frac{dy}{dx}\).

Would you like more details or examples on any part of implicit differentiation?

Here are 5 related questions to deepen your understanding:
1. How is implicit differentiation applied to trigonometric functions?
2. How would you differentiate \(x^2 + xy + y^2 = 10\)?
3. What is the difference between explicit and implicit differentiation?
4. Can implicit differentiation be used to find second derivatives?
5. How does implicit differentiation apply to parametric equations?

**Tip**: Always remember to apply the chain rule when differentiating terms involving \(y\), since \(y\) is treated as a function of \(x\).

Discover how implicit differentiation is used to find the derivative when variables are not easily isolated. This explanation covers key calculus concepts like the chain rule and offers step-by-step examples, such as differentiating \(x^2 + y^2 = 25\). Ideal for students in Grades 11-12.

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