if `y=sqrt(x) + 1/sqrt(x)`, then `(dy)/(dx)` at `x=1` is equal to

Options

A. 1
B. `1/2`
C. `1/sqrt(2)`
D. 0

Explanation

To find \(\frac{dy}{dx}\) for the function \(y = \sqrt{x} + \frac{1}{\sqrt{x}}\) at \(x = 1\), we will follow these steps:

### Step 1: Rewrite the function
The function can be rewritten using exponent notation:
\[ y = x^{1/2} + x^{-1/2} \]

### Step 2: Differentiate the function
Now, we will differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{d}{dx}(x^{1/2}) + \frac{d}{dx}(x^{-1/2}) \]

Using the power rule \(\frac{d}{dx}(x^n) = n \cdot x^{n-1}\), we differentiate each term:
1. For \(x^{1/2}\):
\[ \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} \]
2. For \(x^{-1/2}\):
\[ \frac{d}{dx}(x^{-1/2}) = -\frac{1}{2}x^{-3/2} \]

Combining these results, we have:
\[ \frac{dy}{dx} = \frac{1}{2}x^{-1/2} - \frac{1}{2}x^{-3/2} \]

### Step 3: Substitute \(x = 1\)
Now, we will evaluate \(\frac{dy}{dx}\) at \(x = 1\):
\[ \frac{dy}{dx} \bigg|_{x=1} = \frac{1}{2}(1)^{-1/2} - \frac{1}{2}(1)^{-3/2} \]

Calculating each term:
- \((1)^{-1/2} = 1\)
- \((1)^{-3/2} = 1\)

Thus, we have:
\[ \frac{dy}{dx} \bigg|_{x=1} = \frac{1}{2}(1) - \frac{1}{2}(1) = \frac{1}{2} - \frac{1}{2} = 0 \]

### Final Answer
\[ \frac{dy}{dx} \text{ at } x = 1 \text{ is } 0 \]

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