If f(x)`=4x^2` and g(x) =f(sin x)+f(cos x), then g`(23^(@))` is

Options

A. 1
B. 4
C. 4.29
D. 5.37

Explanation

To solve the problem, we need to find the value of \( g'(23^\circ) \) given that \( f(x) = 4x^2 \) and \( g(x) = f(\sin x) + f(\cos x) \).

### Step-by-Step Solution:

1. **Define the functions**:
We know that:
\[ f(x) = 4x^2 \]
and
\[ g(x) = f(\sin x) + f(\cos x) \]

2. **Substitute \( f(x) \) into \( g(x) \)**:
Using the definition of \( f(x) \):
\[ g(x) = f(\sin x) + f(\cos x) = 4(\sin x)^2 + 4(\cos x)^2 \]

3. **Factor out the common term**:
We can factor out the 4:
\[ g(x) = 4\left((\sin x)^2 + (\cos x)^2\right) \]

4. **Use the Pythagorean identity**:
We know from trigonometry that:
\[ (\sin x)^2 + (\cos x)^2 = 1 \]
Therefore:
\[ g(x) = 4 \cdot 1 = 4 \]

5. **Determine the derivative \( g'(x) \)**:
Since \( g(x) = 4 \) is a constant function, its derivative is:
\[ g'(x) = 0 \]

6. **Evaluate \( g'(23^\circ) \)**:
Thus, we find:
\[ g'(23^\circ) = 0 \]

### Final Answer:
\[ g'(23^\circ) = 0 \]