find the deratives of sin(pi +xlnx). i couldnt find the corredt answer.?
回答 (3)
2015-05-12 3:44 pm
✔ 最佳答案
d/dx sin(pi + x ln x)Let u = pi + x ln x
du/dx = 0 + ln x + x (1/x)
du/dx = ln x + 1
sin ( pi + x ln x) = sin(u)
d/dx sin (pi + x ln x) = d/du ( sin(u)) du/dx
= cos(u) (du/dx)
= cos( pi + x ln x) ( 1n x+ 1)
2015-05-12 3:53 pm
sin(pi + t) =>
sin(pi)cos(t) + sin(t)cos(pi) =>
0 * cos(t) + sin(t) * (-1) =>
-sin(t)
sin(pi + x * ln(x)) =>
-sin(x * ln(x))
u = x * ln(x)
du = x * (1/x) + ln(x) * 1 = 1 + ln(x)
d/du -sin(u) = -cos(u) * du
-cos(x * ln(x)) * (1 + ln(x))
It's the same as the other answers, with slightly less filler material.
sin(pi)cos(t) + sin(t)cos(pi) =>
0 * cos(t) + sin(t) * (-1) =>
-sin(t)
sin(pi + x * ln(x)) =>
-sin(x * ln(x))
u = x * ln(x)
du = x * (1/x) + ln(x) * 1 = 1 + ln(x)
d/du -sin(u) = -cos(u) * du
-cos(x * ln(x)) * (1 + ln(x))
It's the same as the other answers, with slightly less filler material.
2015-05-12 3:40 pm
I got cos(π+x·ln(x))(ln(x)+1)
收錄日期: 2021-04-21 01:48:13
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