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(B) Given the functional equation $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in R$. Setting $x=0$ and $y=0$,we get $f(0)=f(0)^2$,which implies $f(0)(f(0)

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$12$

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(B) Given the functional equation $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in R$. Setting $x=0$ and $y=0$,we get $f(0)=f(0)^2$,which implies $f(0)(f(0)-1)=0$. Since $f(x) 
eq 0$ for all $x$,we must have $f(0)=1$. By the definition of the derivative at $x=0$,$f^{\prime}(0) = \lim_{h ightarrow 0} \frac{f(0+h)-f(0)}{h} = \lim_{h ightarrow 0} \frac{f(h)-1}{h} = 4$. Now,for any $x$,the derivative $f^{\prime}(x)$ is given by: $f^{\prime}(x) = \lim_{h ightarrow 0} \frac{f(x+h)-f(x)}{h} = \lim_{h ightarrow 0} \frac{f(x)f(h)-f(x)}{h} = f(x) \lim_{h ightarrow 0} \frac{f(h)-1}{h}$. Substituting the value of the limit,we get $f^{\prime}(x) = f(x) \cdot 4 = 4f(x)$. Therefore,$f^{\prime}(6) = 4f(6) = 4 	imes 3 = 12$.

$A$ function $f: R \rightarrow R$ satisfies the relation $f(x+y)=f(x) \cdot f(y), \forall x, y \in R$ and $f(x) \neq 0, \forall x \in R$. If $f$ is differentiable at $x=0$,$f^{\prime}(0)=4$,and $f(6)=3$,then $f^{\prime}(6)$ is equal to

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