Question

Let a function $f:R\to R$ be given by $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,y\in R$ and $f\left(x\right)\ne 0$ for any $x\in R$. If the function $f\left(x\right)$ is differentiable at $x=0$, show that $f^{\prime }\left(x\right)=f^{\prime }\left(0\right)f\left(x\right)\text{for all}x\in R$. Also, determine $f\left(x\right)$.

Answer 1

Differentiate wrt x $\to$ $f(x+y)=f\left(x\right)f\left(y\right),xϵR,yϵR$

$f^{\prime }(x+y)=f^{\prime }\left(x\right)f\left(y\right)+f\left(x\right)f^{\prime }\left(y\right)\frac{dy}{dx}$

Put $y=const=0$

$\Rightarrow f^{\prime }\left(x\right)=f^{\prime }\left(x\right)f\left(0\right)+0$

$\Rightarrow f\left(0\right)=1(\because f^{\prime }(x)\ne 0\forall x)$

$f^{\prime }(x+y)=f\left(x\right)f\left(y\right)$

$f^{\prime }(x+y)=f^{\prime }\left(x\right)f\left(y\right)+fl\left(y\right)y^{\prime }.f\left(x\right)$

Let $y=const\Rightarrow y^{\prime }=0$

$\Rightarrow f^{\prime }(x+y)=f^{\prime }\left(x\right)f\left(y\right)$

Put $x=0$

$\Rightarrow f^{\prime }\left(y\right)=f^{\prime }\left(0\right)f\left(y\right)$

or $f^{\prime }\left(x\right)=f^{\prime }\left(0\right)f\left(x\right)$

$\frac{df\left(x\right)}{dx}=f^{\prime }\left(0\right)f\left(x\right)$

$\Rightarrow \frac{df\left(x\right)}{f(x}=f^{\prime }\left(0\right)dx$

$\Rightarrow \int \frac{df\left(x\right)}{f\left(x\right)}=\int f^{\prime }\left(0\right)dx$

$\Rightarrow ln\left|p\right(x\left)\right|=f^{\prime }\left(0\right)x+c$

$\Rightarrow f\left(x\right)=k{e}^{f^{\prime }\left(0\right)x}$

$f\left(0\right)=1\Rightarrow k{e}^{f^{\prime }\left(0\right).0}=1\Rightarrow k=1$

$\Rightarrow f\left(x\right)=e^{f^{\prime }\left(0\right)x}$