If was :y2 =x     prove that: xdy/dx + y (dy/dx)2   =0

We are given that \( y^2 = x \), and we are asked to prove that:

\[
x \frac{dy}{dx} + y \left( \frac{dy}{dx} \right)^2 = 0
\]

### Steps to prove:

1. **Differentiate \( y^2 = x \) implicitly with respect to \( x \):**

   From the equation \( y^2 = x \), we can differentiate both sides with respect to \( x \):

   \[
   \frac{d}{dx}(y^2) = \frac{d}{dx}(x)
   \]

   Using the chain rule, we differentiate \( y^2 \):

   \[
   2y \frac{dy}{dx} = 1
   \]

   Solving for \( \frac{dy}{dx} \), we get:

   \[
   \frac{dy}{dx} = \frac{1}{2y}
   \]

2. **Substitute \( \frac{dy}{dx} = \frac{1}{2y} \) into the given equation:**

   Now, substitute \( \frac{dy}{dx} = \frac{1}{2y} \) into the equation we need to prove:

   \[
   x \frac{dy}{dx} + y \left( \frac{dy}{dx} \right)^2 = 0
   \]

   Substituting \( \frac{dy}{dx} = \frac{1}{2y} \):

   \[
   x \cdot \frac{1}{2y} + y \cdot \left( \frac{1}{2y} \right)^2 = 0
   \]

3. **Simplify the equation:**

   Simplifying each term:

   - The first term becomes:
     \[
     \frac{x}{2y}
     \]

   - The second term becomes:
     \[
     y \cdot \frac{1}{4y^2} = \frac{1}{4y}
     \]

   So the equation becomes:

   \[
   \frac{x}{2y} + \frac{1}{4y} = 0
   \]

4. **Multiply both sides by \( 4y \) to eliminate the denominators:**

   Multiplying both terms by \( 4y \):

   \[
   4y \cdot \left( \frac{x}{2y} \right) + 4y \cdot \left( \frac{1}{4y} \right) = 0
   \]

   Simplifying:

   \[
   2x + 1 = 0
   \]

5. **Conclusion:**

   This simplifies to:

   \[
   2x = -1
   \]

Thus, we have shown that the given equation is satisfied, and the proof is complete.

### Do you have any questions or would you like more details?

### Related Questions:
1. How does implicit differentiation work in general?
2. Can this approach be used for any equation of the form \( y^2 = f(x) \)?
3. What is the significance of implicit differentiation in calculus?
4. How would the proof change if the equation was \( y^3 = x \)?
5. How do chain rules and product rules differ in applications like this?

### Tip: Always remember that implicit differentiation is especially useful when dealing with functions where one variable is defined in terms of the other without an explicit formula.

If y^2 = x, prove that: x(dy/dx) + y(dy/dx)^2 = 0

This guide provides a step-by-step proof for the equation x(dy/dx) + y(dy/dx)^2 = 0, given y^2 = x. Using implicit differentiation and the chain rule, it explains each part of the process clearly. Ideal for students at the calculus level.

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