Differentiate the following w.r.t.x. `sqrt(x^(2)+x+1)`

Answer

To differentiate the function \( y = \sqrt{x^2 + x + 1} \) with respect to \( x \), we will follow these steps:

### Step 1: Rewrite the Function
Let \( y = \sqrt{x^2 + x + 1} \). We can rewrite this as:
\[ y = (x^2 + x + 1)^{1/2} \]

### Step 2: Differentiate Using the Chain Rule
To differentiate \( y \) with respect to \( x \), we will use the chain rule. The chain rule states that if \( y = f(g(x)) \), then:
\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]
Here, \( f(u) = u^{1/2} \) and \( g(x) = x^2 + x + 1 \).

First, we differentiate \( f(u) \):
\[ f'(u) = \frac{1}{2} u^{-1/2} = \frac{1}{2\sqrt{u}} \]
Substituting back \( u = g(x) = x^2 + x + 1 \):
\[ f'(g(x)) = \frac{1}{2\sqrt{x^2 + x + 1}} \]

Next, we differentiate \( g(x) \):
\[ g'(x) = \frac{d}{dx}(x^2 + x + 1) = 2x + 1 \]

### Step 3: Apply the Chain Rule
Now, we can combine these results using the chain rule:
\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) = \frac{1}{2\sqrt{x^2 + x + 1}} \cdot (2x + 1) \]

### Step 4: Simplify the Expression
Thus, the derivative is:
\[ \frac{dy}{dx} = \frac{2x + 1}{2\sqrt{x^2 + x + 1}} \]

### Final Answer
The derivative of \( y = \sqrt{x^2 + x + 1} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{2x + 1}{2\sqrt{x^2 + x + 1}} \]

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