高数导数部分题目。图中的两道题。过程写详细点,因为我是初学,所有请不要跳步骤。
2个回答
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(5)
y1=(1/2) (1+x^2)(arctanx)^2
y1' =(1/2)[ (1+x^2). (2arctanx). (1/(1+x^2) + 2x.(arctanx)^2 ]
=(1/2)[ 2arctanx + 2x.(arctanx)^2 ]
= arctanx + x.(arctanx)^2
y2 =-xarctanx
y2' = -x/(1+x^2) - arctanx
y3 = (1/2)ln(1+x^2)
y3' = (1/2) [ 1/(1+x^2)] .(2x)
= x/(1+x^2)
y=y1+y2+y3
y' =y1'+y2'+y3'
=arctanx + x.(arctanx)^2 -x/(1+x^2) - arctanx + x/(1+x^2)
=x.(arctanx)^2
(6)
y1= (1/2)ln[1+e^(2x)]
y1' = (1/2).( 1/[1+e^(2x)] ). (2e^(2x))
= e^(2x)/[1+e^(2x)]
y2=-2√(1-x^2)arcsinx
y2' = -2{ √(1-x^2) . [1/√(1-x^2)] + arcsinx . (1/2)(1-x^2)^(-1/2) .(-2x) }
=-2[1- xarcsinx/√(1-x^2) ]
y3=e^xarctan(e^x)
y3' =e^xarctan(e^x) + e^x . [1/(1+e^(2x)] . e^x
=e^xarctan(e^x) + e^(2x)/(1+e^(2x)
y=y1+y2+y3
y' =e^(2x)/[1+e^(2x)] -2[1- xarcsinx/√(1-x^2) ]
+e^xarctan(e^x) + e^(2x)/(1+e^(2x)
=2e^(2x)/[1+e^(2x)] -2[1- xarcsinx/√(1-x^2) ] +e^xarctan(e^x)
y1=(1/2) (1+x^2)(arctanx)^2
y1' =(1/2)[ (1+x^2). (2arctanx). (1/(1+x^2) + 2x.(arctanx)^2 ]
=(1/2)[ 2arctanx + 2x.(arctanx)^2 ]
= arctanx + x.(arctanx)^2
y2 =-xarctanx
y2' = -x/(1+x^2) - arctanx
y3 = (1/2)ln(1+x^2)
y3' = (1/2) [ 1/(1+x^2)] .(2x)
= x/(1+x^2)
y=y1+y2+y3
y' =y1'+y2'+y3'
=arctanx + x.(arctanx)^2 -x/(1+x^2) - arctanx + x/(1+x^2)
=x.(arctanx)^2
(6)
y1= (1/2)ln[1+e^(2x)]
y1' = (1/2).( 1/[1+e^(2x)] ). (2e^(2x))
= e^(2x)/[1+e^(2x)]
y2=-2√(1-x^2)arcsinx
y2' = -2{ √(1-x^2) . [1/√(1-x^2)] + arcsinx . (1/2)(1-x^2)^(-1/2) .(-2x) }
=-2[1- xarcsinx/√(1-x^2) ]
y3=e^xarctan(e^x)
y3' =e^xarctan(e^x) + e^x . [1/(1+e^(2x)] . e^x
=e^xarctan(e^x) + e^(2x)/(1+e^(2x)
y=y1+y2+y3
y' =e^(2x)/[1+e^(2x)] -2[1- xarcsinx/√(1-x^2) ]
+e^xarctan(e^x) + e^(2x)/(1+e^(2x)
=2e^(2x)/[1+e^(2x)] -2[1- xarcsinx/√(1-x^2) ] +e^xarctan(e^x)
追问
第五题的答案,我的书上写的是[-2x^2arctanx]/(1+x^2)是答案错了吗?
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