Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

A function $f\;:\;R\rightarrow R$ satisfies the equation $f(x+y)=f(x)f(y)$ for all $x,y\in R,f(x)\neq 0.$ Suppose that the function is differentiable at $x=0$ and $f'(0)=2$.Prove that $f'(x)=2f(x).$

Can you answer this question?

1 Answer

0 votes
  • Chain Rule : Suppose a real valued function which is a composite of two function $u$ and $v$ $(i.e$) $f=uov$ then $\large\frac{df}{dx}=\frac{du}{dt}$$\times \large\frac{dt}{dx}$
  • Product rule : $\large\frac{d}{dx}$$(uv)=u\large\frac{d}{dx}$$v+v\large\frac{d}{dx}$$(u)$
Step 1:
Given : $f(x+y)=f(x)f(y)$.
Now $f(x)$ and $f(y)$ are invertible function.
Hence $f(y)=x$
If $y=f(x)$ then $f(f(x))=x$
Now differentiating by applying the chain rule on LHS and product rule on the RHS.
Step 2:
Put $x=0,y=0$ and $f'(0)=2$
Hence proved.
answered Jun 27, 2013 by sreemathi.v
edited Jun 27, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App