Answer is: option4
\( -\frac{2}{3} \)Solution:
To find \( F'(x) \), we differentiate both sides: \[ F'(x) = G'[x + G(x)] \cdot \frac{d}{dx} [x + G(x)] \] Using basic differentiation: \[ \frac{d}{dx} [x + G(x)] = 1 + G'(x) \]
Thus, we get:
\[ F'(x) = G'[x + G(x)] \cdot (1 + G'(x)) \] We substitute \( x = 1 \):- \( G(1) = 3 \), so the inner function becomes \( 1 + 3 = 4 \).
- \( G'(4) = \frac{2}{3} \), so \( G'[4] = \frac{2}{3} \).
- \( G'(1) = -2 \), so \( 1 + G'(1) = 1 - 2 = -1 \).
Thus,
\[ F'(1) = G'[4] \cdot (-1) = \frac{2}{3} \times (-1) = -\frac{2}{3} \] Final Answer: \[ \boxed{-\frac{2}{3}} \]