Question

If \(y = \left\{x + \sqrt{x^2 - a^2}\right\}^n\), then prove that \(\dfrac{dy}{dx} = \dfrac{ny}{\sqrt{x^2 - a^2}}\).

Hint:

Use the chain rule on the power. The derivative of the inner term simplifies to u divided by the square root, which gives n y over the root.

Answer 1

Step 1: Set up the differentiation.
The function is a power of an expression, so we use the chain rule. Let \(u = x + \sqrt{x^2 - a^2}\), so \(y = u^n\).
\[ \frac{dy}{dx} = n u^{n-1} \cdot \frac{du}{dx} \]

Step 2: Find du/dx.
Differentiate \(\sqrt{x^2 - a^2}\) by the chain rule. Its derivative is \(\dfrac{2x}{2\sqrt{x^2 - a^2}}\).
\[ \frac{du}{dx} = 1 + \frac{x}{\sqrt{x^2 - a^2}} \]
Take the common denominator.
\[ \frac{du}{dx} = \frac{\sqrt{x^2 - a^2} + x}{\sqrt{x^2 - a^2}} = \frac{u}{\sqrt{x^2 - a^2}} \]

Step 3: Substitute back.
Put this value of \(\dfrac{du}{dx}\) in the chain rule result.
\[ \frac{dy}{dx} = n u^{n-1} \cdot \frac{u}{\sqrt{x^2 - a^2}} = \frac{n u^{n}}{\sqrt{x^2 - a^2}} \]

Step 4: Replace u^n by y.
Since \(y = u^n\), we get the required result.

Final Answer:
\[ \frac{dy}{dx} = \frac{ny}{\sqrt{x^2 - a^2}} \quad \text{Hence proved.} \]