Let f(x+y)=f(x)f(y) for all x and y, suppose f(5)=2, andf’(0)=3, then f’(5) equals to
6
7
4
8
Step 1: Determine the value of f(0)
Let us take x=5 and y=0. Now:
f(x+y)=f(x)f(y)⇒f(5+0)=f(5)f(0)⇒f(5)=2×f(0)⇒2=2×f(0)⇒f(0)=1
Step 2: Differentiate f(x+y)=f(x)f(y) with respect to x.
f'(x+y)×1+y'=f'(x)f(y)+f(x)f'(y)y'⇒f'(5+0)×(1+y')=f'(5)f(0)+f(5)f'(0)y'⇒f'(5)(1+y')=f'(5)×1+(2×3)y'⇒f'(5)(1+y')=f'(5)+6y'⇒f'(5)(1+y')-f'(5)=6y'⇒f'(5)(1+y'-1)=6y'⇒f'(5)y'=6y'⇒f'(5)=6
Hence, option A is the correct answer.