f(x)=2x9=2x−9f′(x)=−18x−10f′(2)=−18×2−10=−9512f′(−3)=−18×(−3)−10=−238=−26561\displaystyle f(x) = \frac{2}{x^9} = 2x^{-9}\\ f'(x) = -18x^{-10}\\ f'(2) = -18\times 2^{-10} = \frac{-9}{512}\\ f'(-3) = -18\times (-3)^{-10} = \frac{-2}{3^8} = \frac{-2}{6561}f(x)=x92=2x−9f′(x)=−18x−10f′(2)=−18×2−10=512−9f′(−3)=−18×(−3)−10=38−2=6561−2
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