Implicit differentiation is a technique used to differentiate equations where the variables $x$ and $y$ are not explicitly separated, meaning $y$ is not isolated on one side of the equation. Here's an example with a step-by-step explanation:

Example:

Differentiate the equation $x^2 + y^2 = 25$ implicitly with respect to $x$.

Step-by-step solution:

1. Differentiate both sides with respect to $x$: Since $y$ is a function of $x$, when differentiating any term involving $y$, we apply the chain rule.

   $\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25)$

2. Apply the derivative:

   • The derivative of $x^2$ with respect to $x$ is $2x$.
   • The derivative of $y^2$ with respect to $x$ is $2y \frac{dy}{dx}$ by the chain rule (since $y$ is a function of $x$).
   • The derivative of the constant $25$ is $0$.

   Thus, the equation becomes: $2x + 2y \frac{dy}{dx} = 0$

3. Solve for $\frac{dy}{dx}$: Isolate $\frac{dy}{dx}$ by moving the terms around:

   $2y \frac{dy}{dx} = -2x$

   Divide both sides by $2y$:

   $\frac{dy}{dx} = \frac{-x}{y}$

Final Answer:

The derivative $\frac{dy}{dx}$ is $\frac{-x}{y}$.

This shows how to implicitly differentiate when the variables are mixed in an equation.

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Would you like more details on any of the steps? Here are some related questions:

1. How do you apply the chain rule in implicit differentiation?
2. What if the equation involved higher powers of $x$ and $y$?
3. Can implicit differentiation be used with trigonometric functions?
4. How do you apply implicit differentiation to find the second derivative?
5. What happens if you need to solve for $y$ after differentiating?

Tip: When performing implicit differentiation, always apply the chain rule to terms involving $y$ since $y$ is treated as a function of $x$.