(3/2x-1/2)^9=a0+a1x+a2x^+……a8x^+a9x^9,则a1+2a2+3a3+……8a8+9a9 =
人气:227 ℃ 时间:2020-06-29 05:16:13
解答
(a0+a1x+a2x^2+……a8x^+a9x^9)′=(a1+2a2x+3a3x^2+……8a8x^7+9a9x^8)f′(1)=a1+2a2+3a3+……8a8+9a9=(3/2x-1/2)^9′=9(3/2x-1/2)^8*3/2=9*1*3/2=27/2【数学辅导团】为您解答,如果本题有什么不明白可以追问,...=(3/2x-1/2)^9′ =9(3/2x-1/2)^8*3/2为什么9(3/2x-1/2)^8后面还要乘3/2?复合函数求导数,还要对(3/2x-1/2)′求一次导,是3/2啊我明白了还得乘以内导非常感谢!!!!
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