Find the derivative of `y=sqrt(x^2+1)`.

Options

A. `(4x)/(sqrt(x^2-1))`
B. `(2x)/(sqrt(2x^2+1))`
C. `(x)/(sqrt(x^2-1))`
D. `(x)/(sqrt(x^2+1))`

Explanation

To find the derivative of the function \( y = \sqrt{x^2 + 1} \), we will use the chain rule of differentiation. Here’s a step-by-step solution:

### Step 1: Rewrite the function
We start with the function:
\[ y = \sqrt{x^2 + 1} \]
This can be rewritten using exponent notation:
\[ y = (x^2 + 1)^{1/2} \]

### Step 2: Apply the chain rule
To differentiate \( y \) with respect to \( x \), we will apply the chain rule. The chain rule states that if \( y = f(g(x)) \), then:
\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]
In our case, let \( g(x) = x^2 + 1 \) and \( f(g) = g^{1/2} \).

### Step 3: Differentiate the outer function
First, we differentiate the outer function \( f(g) = g^{1/2} \):
\[ f'(g) = \frac{1}{2} g^{-1/2} \]
Now substituting back \( g(x) \):
\[ f'(g(x)) = \frac{1}{2} (x^2 + 1)^{-1/2} \]

### Step 4: Differentiate the inner function
Next, we differentiate the inner function \( g(x) = x^2 + 1 \):
\[ g'(x) = 2x \]

### Step 5: Combine the results
Now we can combine the results from the chain rule:
\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) = \frac{1}{2} (x^2 + 1)^{-1/2} \cdot 2x \]
Simplifying this gives:
\[ \frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 1}} \]

### Final Answer
Thus, the derivative of \( y = \sqrt{x^2 + 1} \) is:
\[ \frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 1}} \]

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