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If $f$ is differentiable, $f(x+y)=f(x) f(y)$ for all $x, y \in \mathbb{R}, f(3)=3, f^{\prime}(0)=11$, then $f^{\prime}(3)=$
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Verified Answer
The correct answer is:
33
We have,
$f(x+y)=f(x) f(y)$
differentiate with respect to $x, y$ as constant, we get
$$
f^{\prime}(x+y)=f^{\prime}(x) f(y)
$$
Put $x=0, y=3$, we get
$$
\begin{aligned}
f^{\prime}(0+3) & =f^{\prime}(0) f(3) & \\
\Rightarrow \quad f^{\prime}(3) & =f^{\prime}(0) \cdot f(3) & \\
\Rightarrow f^{\prime}(3)=11 \times 3 & =33 & {\left[\because f^{\prime}(0)=11, f(3)=3\right] }
\end{aligned}
$$
$f(x+y)=f(x) f(y)$
differentiate with respect to $x, y$ as constant, we get
$$
f^{\prime}(x+y)=f^{\prime}(x) f(y)
$$
Put $x=0, y=3$, we get
$$
\begin{aligned}
f^{\prime}(0+3) & =f^{\prime}(0) f(3) & \\
\Rightarrow \quad f^{\prime}(3) & =f^{\prime}(0) \cdot f(3) & \\
\Rightarrow f^{\prime}(3)=11 \times 3 & =33 & {\left[\because f^{\prime}(0)=11, f(3)=3\right] }
\end{aligned}
$$
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