赞 诽色拓己 幼苗
共回答了13个问题采纳率：84.6% 举报
y = tan(x + y)
y' = (x+y)'/cos²(x+y) = (1 + y')/cos²(x+y)
y'cos²(x+y) = 1 + y'
[cos²(x+y) - 1]y' = 1
[cos²(x+y) - sin²(x+y) - cos²(x+y)]y' = 1
y' = -1/sin²(x+y)
y'' = -1*(-2)[sin(x+y)]'/sin³(x+y)
= 2cos(x+y)(x+y)'/sin³(x+y)
= 2cos(x+y)(1 + y')/sin³(x+y)
= 2cos(x+y)[1 -1/sin²(x+y)]/sin³(x+y)
= 2cos(x+y)[sin²(x+y) -1]/[sin(x+y)]^5
= 2cos(x+y)[sin²(x+y) - sin²(x+y) - cos²(x+y)]/[sin(x+y)]^5
= -2cos³(x+y)/[sin(x+y)]^5 (^5:5次方)
y' = (x+y)'/cos²(x+y) = (1 + y')/cos²(x+y)
y'cos²(x+y) = 1 + y'
[cos²(x+y) - 1]y' = 1
[cos²(x+y) - sin²(x+y) - cos²(x+y)]y' = 1
y' = -1/sin²(x+y)
y'' = -1*(-2)[sin(x+y)]'/sin³(x+y)
= 2cos(x+y)(x+y)'/sin³(x+y)
= 2cos(x+y)(1 + y')/sin³(x+y)
= 2cos(x+y)[1 -1/sin²(x+y)]/sin³(x+y)
= 2cos(x+y)[sin²(x+y) -1]/[sin(x+y)]^5
= 2cos(x+y)[sin²(x+y) - sin²(x+y) - cos²(x+y)]/[sin(x+y)]^5
= -2cos³(x+y)/[sin(x+y)]^5 (^5:5次方)
1年前
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