if `y= {x+ sqrt""(x^2 + a^2 )} ^n ` then ` (dy)/(dx)` =

Options

A. `(nx)/(sqrt""(x^2 +a^2))`
B. `(n)/(sqrt""(x^2 +a^2))`
C. `(ny)/(sqrt""(x^2+a^2))`
D. none

Answer

To find the derivative of the function \( y = (x + \sqrt{x^2 + a^2})^n \), we will apply the chain rule for differentiation. Here’s a step-by-step solution:

### Step 1: Identify the outer and inner functions
We can express the function as:
- Outer function: \( u^n \) where \( u = x + \sqrt{x^2 + a^2} \)
- Inner function: \( u = x + \sqrt{x^2 + a^2} \)

### Step 2: Differentiate the outer function
Using the chain rule, the derivative of \( y \) with respect to \( x \) is given by:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]
The derivative of the outer function \( u^n \) is:
\[ \frac{dy}{du} = n \cdot u^{n-1} \]

### Step 3: Differentiate the inner function
Now, we need to find \( \frac{du}{dx} \):
\[ u = x + \sqrt{x^2 + a^2} \]
To differentiate \( u \), we will differentiate each term:
1. The derivative of \( x \) is \( 1 \).
2. For \( \sqrt{x^2 + a^2} \), we use the chain rule:
\[ \frac{d}{dx}(\sqrt{x^2 + a^2}) = \frac{1}{2\sqrt{x^2 + a^2}} \cdot \frac{d}{dx}(x^2 + a^2) = \frac{1}{2\sqrt{x^2 + a^2}} \cdot (2x) = \frac{x}{\sqrt{x^2 + a^2}} \]

So, we have:
\[ \frac{du}{dx} = 1 + \frac{x}{\sqrt{x^2 + a^2}} \]

### Step 4: Combine the derivatives
Now, substituting back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = n \cdot (x + \sqrt{x^2 + a^2})^{n-1} \cdot \left(1 + \frac{x}{\sqrt{x^2 + a^2}}\right) \]

### Step 5: Simplify the expression
We can simplify \( 1 + \frac{x}{\sqrt{x^2 + a^2}} \):
\[ 1 + \frac{x}{\sqrt{x^2 + a^2}} = \frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}} \]

Thus, the derivative becomes:
\[ \frac{dy}{dx} = n \cdot (x + \sqrt{x^2 + a^2})^{n-1} \cdot \frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}} \]

### Final Result
The final expression for the derivative is:
\[ \frac{dy}{dx} = n \cdot (x + \sqrt{x^2 + a^2})^{n-1} \cdot \frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}} \]

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