Find `dy/dx` for each of the following
`y=(sqrtx+1/sqrtx)(1+x+x^(2))`

Answer

To find \(\frac{dy}{dx}\) for the function

\[ y = \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)(1 + x + x^2), \]

we will use the product rule of differentiation, which states that if \(y = u \cdot v\), then

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}. \]

### Step 1: Identify \(u\) and \(v\)

Let
\[ u = \sqrt{x} + \frac{1}{\sqrt{x}} \quad \text{and} \quad v = 1 + x + x^2. \]

### Step 2: Differentiate \(u\)

To differentiate \(u\), we need to find \(\frac{du}{dx}\):

\[ u = \sqrt{x} + \frac{1}{\sqrt{x}} = x^{1/2} + x^{-1/2}. \]

Now, differentiate \(u\):

\[ \frac{du}{dx} = \frac{1}{2}x^{-1/2} - \frac{1}{2}x^{-3/2} = \frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}}. \]

### Step 3: Differentiate \(v\)

Next, differentiate \(v\):

\[ v = 1 + x + x^2. \]

Now, differentiate \(v\):

\[ \frac{dv}{dx} = 0 + 1 + 2x = 1 + 2x. \]

### Step 4: Apply the Product Rule

Now we can apply the product rule:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}. \]

Substituting the values we found:

\[ \frac{dy}{dx} = \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)(1 + 2x) + (1 + x + x^2)\left(\frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}}\right). \]

### Step 5: Simplify the Expression

Now, we will simplify the expression:

1. Expand the first term:
\[ \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)(1 + 2x) = \sqrt{x} + 2x\sqrt{x} + \frac{1}{\sqrt{x}} + \frac{2x}{\sqrt{x}} = \sqrt{x} + 2x\sqrt{x} + x^{-1/2} + 2x^{1/2}. \]

2. Expand the second term:
\[ (1 + x + x^2)\left(\frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}}\right) = \frac{1}{2\sqrt{x}} + \frac{x}{2\sqrt{x}} + \frac{x^2}{2\sqrt{x}} - \left(\frac{1}{2x^{3/2}} + \frac{x}{2x^{3/2}} + \frac{x^2}{2x^{3/2}}\right). \]

Combining these terms will give us the final expression for \(\frac{dy}{dx}\).

### Final Result

Thus, the derivative \(\frac{dy}{dx}\) can be expressed as:

\[ \frac{dy}{dx} = \left(\sqrt{x} + 2x\sqrt{x} + \frac{1}{\sqrt{x}} + 2x^{1/2}\right) + \left(\frac{1 + x + x^2}{2\sqrt{x}} - \frac{1 + x + x^2}{2x^{3/2}}\right). \]