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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 $f'(x)=2f(x).$

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Toolbox:
  • 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.
$f'(x+y)(1+f'(x))=f(x).f'(y)+f(y).f'(x)$
Step 2:
Put $x=0,y=0$ and $f'(0)=2$
$f'(x)=f(x)f'(0)+x.f(x)$
$f'(x)=2f(x)$
Hence proved.
answered Jun 27, 2013 by sreemathi.v
edited Jun 27, 2013 by sreemathi.v
 

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