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\mathrm{f}^{\prime}(\mathrm{x})=\lim\limits_{h \rightarrow 0} \frac{\left\{2(x+h)^{3}-5(x+h)\right\}-\left\{2 x^{3}-5 x\right\}}{h}=\lim\limits_{h \rightarrow 0} \frac{6 x^{2} h+6 x h^{2}+2 h^{3}-5 h}{h}=6 x^{2}-5 . \mathrm{f}^{\prime}(3)=49