We are given the function \( y = \sqrt{x^2 + a^2} \). Our goal is to prove that:

\[
y \frac{dy}{dx} - x = 0
\]

### Step 1: Differentiate \( y \) with respect to \( x \)

We start by differentiating both sides of the equation \( y = \sqrt{x^2 + a^2} \) with respect to \( x \).

\[
y = \sqrt{x^2 + a^2}
\]

Now, applying the chain rule:

\[
\frac{dy}{dx} = \frac{1}{2} (x^2 + a^2)^{-\frac{1}{2}} \cdot 2x
\]

Simplifying:

\[
\frac{dy}{dx} = \frac{x}{\sqrt{x^2 + a^2}}
\]

### Step 2: Substitute into the given equation

The given equation is:

\[
y \frac{dy}{dx} - x = 0
\]

Substitute \( y = \sqrt{x^2 + a^2} \) and \( \frac{dy}{dx} = \frac{x}{\sqrt{x^2 + a^2}} \) into the equation:

\[
\sqrt{x^2 + a^2} \cdot \frac{x}{\sqrt{x^2 + a^2}} - x = 0
\]

Simplifying the left-hand side:

\[
\frac{x \sqrt{x^2 + a^2}}{\sqrt{x^2 + a^2}} - x = 0
\]

The square roots cancel out:

\[
x - x = 0
\]

Thus, the equation holds true.

### Conclusion

We have successfully proven that:

\[
y \frac{dy}{dx} - x = 0
\]

Do you want more details or have any questions?

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Here are 5 related questions:

1. How do you apply the chain rule to more complex functions?
2. What are the applications of this differentiation technique in physics?
3. Can you prove the same result using implicit differentiation?
4. How would this proof change if \( a \) were a function of \( x \)?
5. What happens if you differentiate this result again with respect to \( x \)?

**Tip**: Whenever differentiating a square root, use the chain rule and simplify step by step for accuracy.

If y = √(x^2 + a^2), prove that y * dy/dx - x = 0

Explore the detailed proof of y = √(x^2 + a^2) and y * dy/dx - x = 0 using the chain rule. This step-by-step solution walks through key differentiation techniques and algebraic simplifications, making it ideal for students in Grades 11-12.

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