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If $f(x)$ is an odd differentiable function defined on $(-\infty, \infty)$ such that $f^{\prime}(3)=2,$ then $f^{\prime}(-3)$ is equal to
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Given that $f(x)$ is an odd differentiable function.
Then, $f(-x)=-f(x)$
$\begin{array}{lr}\Rightarrow & -f^{\prime}(-x)=-f^{\prime}(x) \\ \Rightarrow & f^{\prime}(-x)=f^{\prime}(x)\end{array}$
Put $x=3$ in Eq. (i), we get
$\therefore$
$$
\begin{array}{l}
f^{\prime}(-3)=f^{\prime}(3) \\
f^{\prime}(-3)=2
\end{array}
$$
Then, $f(-x)=-f(x)$
$\begin{array}{lr}\Rightarrow & -f^{\prime}(-x)=-f^{\prime}(x) \\ \Rightarrow & f^{\prime}(-x)=f^{\prime}(x)\end{array}$
Put $x=3$ in Eq. (i), we get
$\therefore$
$$
\begin{array}{l}
f^{\prime}(-3)=f^{\prime}(3) \\
f^{\prime}(-3)=2
\end{array}
$$
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