Concept:Recognize the given limit as f′(0) and apply the product rule for differentiation.Explanation:Given g(x)=(x2+2x+1)⋅f(x).Since x2+2x+1=(x+1)2, we can write g(x)=(x+1)2f(x).Now, limx→0xf(x)−5=4.But f(0)=5, so this limit becomes limx→0xf(x)−f(0)=f′(0).Hence, f′(0)=4.Differentiate g(x) using the product rule:g′(x)=2(x+1)f(x)+(x+1)2f′(x).Substitute x=0:g′(0)=2(0+1)f(0)+(0+1)2f′(0).g′(0)=2(1)(5)+(1)2(4)=10+4=14.Answer:g′(0)=14, so the correct option is D.