Show that
d/dx { x{[(x^2) +1 ] ^n} } = (2n+1) {[ (x^2) +1]^n} - 2n { [(x^2) +1]^(n-1)}
thanks a lot if u can help me
amaths differentiation
2007-11-28 6:15 am
回答 (1)
2007-11-28 6:36 am
✔ 最佳答案
d/dx { x{[(x^2) +1 ] ^n} } = {[(x^2) +1 ] ^n} d(x)/dx + x d{[(x^2) +1 ] ^n} /dx
= {[(x^2) +1 ] ^n} (1) +x{ n[(x^2 + 1)^(n-1)] (2x)}
= {[(x^2) +1 ] ^n} + 2nx^2(x^2 + 1)^(n-1)
= {[(x^2) +1 ] ^n} + 2n(x^2+1)(x^2 + 1)^(n-1) - 2n(x^2 + 1)^(n-1)
= (2n+1){[(x^2) +1 ] ^n} - 2n(x^2 + 1)^(n-1)
= (2n+1) {[ (x^2) +1]^n} - 2n { [(x^2) +1]^(n-1)}
收錄日期: 2021-04-13 18:12:51
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