We are given differentiable functions \( f \) and \( g \) in \( (0,1) \) that satisfy the conditions: \[ f(0) = 2, \quad g(1) = 2, \quad g(0) = 0, \quad f(1) = 6 \] We aim to find a relation involving \( f'(c) \) and \( g'(c) \) for some \( c \in (0,1) \). Step 1: Define a New Function Consider the function: \[ h(x) = 2f(x) - g(x) \] Step 2: Compute \( h(0) \) and \( h(1) \) \[ h(0) = 2f(0) - g(0) = 2(2) - 0 = 4 \] \[ h(1) = 2f(1) - g(1) = 2(6) - 2 = 12 - 2 = 10 \] Since \( h(x) \) is differentiable in \( (0,1) \), by the Mean Value Theorem (MVT), there exists some \( c \in (0,1) \) such that: \[ h'(c) = \frac{h(1) - h(0)}{1 - 0} \] \[ h'(c) = \frac{10 - 4}{1} = 6 \] Step 3: Compute \( h'(x) \) Differentiating \( h(x) \): \[ h'(x) = 2f'(x) - g'(x) \] Setting \( h'(c) = 6 \), we get: \[ 2f'(c) - g'(c) = 6 \] Step 4: Use MVT for \( g(x) \) By MVT, there exists some \( c \in (0,1) \) such that: \[ g'(c) = \frac{g(1) - g(0)}{1 - 0} = \frac{2 - 0}{1} = 2 \] Step 5: Solve for \( f'(c) \) Substituting \( g'(c) = 2 \) into \( 2f'(c) - g'(c) = 6 \): \[ 2f'(c) - 2 = 6 \] \[ 2f'(c) = 8 \] \[ f'(c) = 4 \] Since \( g'(c) = 2 \), we get: \[ f'(c) = 2 g'(c) \] Thus, the correct answer is: \[ \boxed{f'(c) = 2g'(c)} \]