If `f(x)=((3x+1)(2sqrt(x)-1))/(sqrt(x))`, then `f'(1)` is equal to

Options

A. 5
B. `-5`
C. 6
D. `(11)/(2)`

Answer

To find \( f'(1) \) for the function \( f(x) = \frac{(3x + 1)(2\sqrt{x} - 1)}{\sqrt{x}} \), we will follow these steps:

### Step 1: Rewrite the function
We start with the given function:
\[ f(x) = \frac{(3x + 1)(2\sqrt{x} - 1)}{\sqrt{x}} \]
We can rewrite this as:
\[ f(x) = (3x + 1)(2\sqrt{x} - 1) \cdot x^{-1/2} \]

### Step 2: Differentiate the function
We will use the product rule to differentiate \( f(x) \). The product rule states that if \( u(x) \) and \( v(x) \) are two functions, then:
\[ (uv)' = u'v + uv' \]
Let:
\[ u(x) = (3x + 1)(2\sqrt{x} - 1) \quad \text{and} \quad v(x) = x^{-1/2} \]

First, we need to find \( u'(x) \):
\[ u(x) = (3x + 1)(2\sqrt{x} - 1) \]
Using the product rule again on \( u(x) \):
\[ u'(x) = (3)(2\sqrt{x} - 1) + (3x + 1)\left(\frac{d}{dx}(2\sqrt{x} - 1)\right) \]
Now, differentiate \( 2\sqrt{x} - 1 \):
\[ \frac{d}{dx}(2\sqrt{x}) = 2 \cdot \frac{1}{2} x^{-1/2} = x^{-1/2} \]
So,
\[ u'(x) = 3(2\sqrt{x} - 1) + (3x + 1)(x^{-1/2}) \]

Next, we differentiate \( v(x) \):
\[ v'(x) = -\frac{1}{2} x^{-3/2} \]

Now we can apply the product rule:
\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]
Substituting \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \):
\[ f'(x) = \left[3(2\sqrt{x} - 1) + (3x + 1)x^{-1/2}\right] x^{-1/2} + (3x + 1)(2\sqrt{x} - 1)\left(-\frac{1}{2} x^{-3/2}\right) \]

### Step 3: Evaluate \( f'(1) \)
Now we will evaluate \( f'(1) \):
1. Calculate \( u(1) \):
\[ u(1) = (3(1) + 1)(2\sqrt{1} - 1) = 4 \cdot 1 = 4 \]
2. Calculate \( u'(1) \):
\[ u'(1) = 3(2(1) - 1) + (3(1) + 1)(1^{-1/2}) = 3(1) + 4 = 3 + 4 = 7 \]
3. Calculate \( v(1) \):
\[ v(1) = 1^{-1/2} = 1 \]
4. Calculate \( v'(1) \):
\[ v'(1) = -\frac{1}{2}(1^{-3/2}) = -\frac{1}{2} \]

Now substitute these values into the derivative:
\[ f'(1) = (7)(1) + (4)\left(-\frac{1}{2}\right) = 7 - 2 = 5 \]

### Final Answer
Thus, \( f'(1) = 5 \).