Let ϕ(x)=f(x)−2g(x),x∈[0,1] Clearly, ϕ(x) is continuous on [0, 1] and differentiable on (0,1), as f(x) and g(x) are differentiable on [0,1]. Also, ϕ(0)=f(0)−2g(0)=2−0=2 and ϕ(1)=f(1)−2g(1)=6−2×2=2 ∵ ϕ(0)=ϕ(1) Thus, ϕ(x) satisfies all the three conditions of Rolle's theorem. Therefore, there exists a point x∈(0,1) such that ϕ′(x)=0⇒f′(x)−2g′(x)=0 ∴ f′(x)=2g′(x)