If f is differentiable, f(x+y)=f(x) f(y) for | vedclass

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(D) Given the functional equation $f(x+y)=f(x) f(y)$.By the definition of the derivative, $f^{\prime}(x) = \lim_{h 	o 0} \frac{f(x+h)-f(x)}{h}$.Using

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$33$

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(D) Given the functional equation $f(x+y)=f(x) f(y)$.By the definition of the derivative, $f^{\prime}(x) = \lim_{h 	o 0} \frac{f(x+h)-f(x)}{h}$.Using the given relation, $f(x+h) = f(x)f(h)$.So, $f^{\prime}(x) = \lim_{h 	o 0} \frac{f(x)f(h)-f(x)}{h} = f(x) \lim_{h 	o 0} \frac{f(h)-1}{h}$.Since $f(0+0) = f(0)f(0)$, we have $f(0) = f(0)^2$, so $f(0)=1$ (assuming $f(x) 
eq 0$).Thus, $f^{\prime}(0) = \lim_{h 	o 0} \frac{f(h)-f(0)}{h} = \lim_{h 	o 0} \frac{f(h)-1}{h} = 11$.Therefore, $f^{\prime}(x) = f(x) \cdot 11$.At $x=3$, $f^{\prime}(3) = f(3) \cdot 11 = 3 \cdot 11 = 33$.

If $f$ is differentiable, $f(x+y)=f(x) f(y)$ for all $x, y \in R$, $f(3)=3$, and $f^{\prime}(0)=11$, then $f^{\prime}(3)$ is equal to:

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