If `f(x)=x^(2)g (x) and g(1)=6, g'(x) 3`, find the value of f' (1).

Answer

To find the value of \( f'(1) \) for the function \( f(x) = x^2 g(x) \), we will apply the product rule of differentiation. Here are the steps to solve the problem:

### Step 1: Identify the functions
We have:
- \( f(x) = x^2 g(x) \)

### Step 2: Apply the product rule
The product rule states that if you have two functions \( u(x) \) and \( v(x) \), then the derivative \( (uv)' = u'v + uv' \). In our case:
- Let \( u(x) = x^2 \) and \( v(x) = g(x) \)

The derivatives are:
- \( u'(x) = 2x \)
- \( v'(x) = g'(x) \)

Using the product rule, we get:
\[ f'(x) = u'(x)v(x) + u(x)v'(x) = (2x)g(x) + (x^2)g'(x) \]

### Step 3: Substitute known values
Now we need to evaluate \( f'(1) \):
\[ f'(1) = 2(1)g(1) + (1^2)g'(1) \]
We know from the problem statement:
- \( g(1) = 6 \)
- \( g'(1) = 3 \)

Substituting these values in:
\[ f'(1) = 2(1)(6) + (1)(3) \]
\[ f'(1) = 12 + 3 \]
\[ f'(1) = 15 \]

### Final Answer
Thus, the value of \( f'(1) \) is \( 15 \).

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