Given, f(5)=2 and f′(0)=3
f(x+y)=f(x)⋅f(y) ...(1)
Put x=5 and y=0, we get
f(5)=f(5)⋅f(0)
⇒f(0)=1
Differentiating (1) w.r.t x
f′(x+y)(1+y′)=f′(x)⋅f(y)+f(x)⋅f′(y)⋅y′
Put x=5 and y=0, we get
⇒f′(5)(1+y′)=f′(5)⋅f(0)+f(5)⋅f′(0)⋅y′
Since, f(0)=1, f(5)=2 and f′(0)=3
∴f′(5)+y′f′(5)=f′(5)+2⋅3⋅y′
⇒y′f′(5)=6y′
⇒f′(5)=6