作业帮y=ln根号下(1-x)^e^x/arccosx求导
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y=ln根号下(1-x)^e^x/arccosx求导
原式这样
原式这样
y = ln√(1-x)^(e^x)/ arccosx
u = ln√(1-x)^(e^x) = ln (1-x)^[(1/2)e^x]
u' = [1/(1-x)^{(1/2)(e^x)}] .{ ((1/2)e^x) (1-x)^[(1/2)e^x -1] } [ (1/2)e^x ] (-1)
= -e^(2x)/[4(1-x)]
v= arccosx
cosv =x
-sinv v' =1
v' = -1/√(1-x^2)
y = u/v
y' =(vdu-udv)/u^2
=[(arccosx)(-e^(2x)/[4(1-x)]) - (ln√(1-x)^(e^x))(-1/√(1-x^2))] /(arcosx)^2
= { -(arccosx)e^(2x)/[4(1-x)] + ln√(1-x)^(e^x) /√(1-x^2)] /(arcosx)^2
再问： 这种形式真的看着太费劲了
再答： y = ln√[(1-x)^(e^x)/ arccosx] = (1/2)ln[(1-x)^(e^x)/ arccosx] u = (1-x)^(e^x) u' = (e^x)(1-x)^(e^x -1) (e^x) (-1) = -e^(2x)(1-x)^(e^x -1) v =arccosx cosv=x -sinv v' = 1 v' = -1/√(1-x^2) y = (1/2)ln(u/v) y' = (1/2) (v/u) [vu'-uv']/v^2 = (1/2)[ arccosx/(1-x)^(e^x)] [(arccosx)(-e^(2x)(1-x)^(e^x -1))+ (1-x)^(e^x)(1/√(1-x^2)) /(arcosx)^2] = (1/2) {1/[(arcosx)(1-x)^(e^x)]} . [ (1-x)^(e^x)/√(1-x^2) - (e^(2x)(1-x)^(e^x -1))(arccosx)]
u = ln√(1-x)^(e^x) = ln (1-x)^[(1/2)e^x]
u' = [1/(1-x)^{(1/2)(e^x)}] .{ ((1/2)e^x) (1-x)^[(1/2)e^x -1] } [ (1/2)e^x ] (-1)
= -e^(2x)/[4(1-x)]
v= arccosx
cosv =x
-sinv v' =1
v' = -1/√(1-x^2)
y = u/v
y' =(vdu-udv)/u^2
=[(arccosx)(-e^(2x)/[4(1-x)]) - (ln√(1-x)^(e^x))(-1/√(1-x^2))] /(arcosx)^2
= { -(arccosx)e^(2x)/[4(1-x)] + ln√(1-x)^(e^x) /√(1-x^2)] /(arcosx)^2
再问： 这种形式真的看着太费劲了
再答： y = ln√[(1-x)^(e^x)/ arccosx] = (1/2)ln[(1-x)^(e^x)/ arccosx] u = (1-x)^(e^x) u' = (e^x)(1-x)^(e^x -1) (e^x) (-1) = -e^(2x)(1-x)^(e^x -1) v =arccosx cosv=x -sinv v' = 1 v' = -1/√(1-x^2) y = (1/2)ln(u/v) y' = (1/2) (v/u) [vu'-uv']/v^2 = (1/2)[ arccosx/(1-x)^(e^x)] [(arccosx)(-e^(2x)(1-x)^(e^x -1))+ (1-x)^(e^x)(1/√(1-x^2)) /(arcosx)^2] = (1/2) {1/[(arcosx)(1-x)^(e^x)]} . [ (1-x)^(e^x)/√(1-x^2) - (e^(2x)(1-x)^(e^x -1))(arccosx)]