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If $f(x)$ and $g(x)$ are two functions with $g(x)=x-\frac{1}{x}$ and $f \circ g(x)=x^3-\frac{1}{x^3}$, then $f^{\prime}(x)$ is equals to
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The correct answer is:
$3 x^2+3$
Here, $g(x)=x-\frac{1}{x}$
$\begin{aligned} & f \circ g(x)=x^3-\frac{1}{x^3}=\left(x-\frac{1}{x}\right)^3+3 x \cdot \frac{1}{x}\left(x-\frac{1}{x}\right) \\ & f \circ g(x)=\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right) \\ & \therefore f(x)=x^3+3 x \\ & f^{\prime}(x)=3 x^2+3\end{aligned}$
$\begin{aligned} & f \circ g(x)=x^3-\frac{1}{x^3}=\left(x-\frac{1}{x}\right)^3+3 x \cdot \frac{1}{x}\left(x-\frac{1}{x}\right) \\ & f \circ g(x)=\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right) \\ & \therefore f(x)=x^3+3 x \\ & f^{\prime}(x)=3 x^2+3\end{aligned}$
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