dx loge(f′(x))=loge(f(x))+c......(1) since f(0)=1 and f′(0)=2 then c=loge2 Substituting the value of c in equation (1), we get loge(f′(x))=loge(f(x))+loge2 f′(x)=2f(x)
f′(x)
f(x)
=2 Integrating both sides w. r. t. x ∫
f′(x)
f(x)
dx=∫2dx logef(x)=2x+c For f(0)=1,c=0 This gives, logef(x)=2x f(x)=e2x