若当 $\lim _{\Delta x \rightarrow 0} \frac{f\left(x_{0}\right)-f\left(x_{0}+3 \Delta x\right)}{2 \Delta x}=1$, 则 $\mathrm{f}^{\prime}\left(x_{0}\right)$ 等于 $(\quad)$.
解析: 解答: $\lim _{\Delta x \rightarrow 0} \frac{f\left(x_{0}\right)-f\left(x_{0}+3 \Delta x\right)}{2 \Delta x}=-\lim _{\Delta x \rightarrow 0}\left(\frac{f\left(x_{0}+3 \Delta x\right)-f\left(x_{0}\right)}{3 \Delta x} \times \frac{3}{2}\right)$

$=-\frac{3}{2} \lim _{\Delta x \rightarrow 0} \frac{f\left(x_{0}+3 \Delta x\right)-f\left(x_{0}\right)}{3 \Delta x}=-\frac{3}{2} \mathrm{f}^{\prime}\left(x_{0}\right)$.

$\therefore-\frac{3}{2} \mathrm{f}^{\prime}\left(x_{0}\right)=1, \quad \therefore \mathrm{f}^{\prime}\left(x_{0}\right)=-\frac{2}{3}$.

分析: 本题主要考查了导数的概念, 解决问题的关键是根据所给条件结合导数的概念进行计算即可.