Question

求下列函数的导数。

20，21，22这三题的过程

Answer 1

方法如下，
请作参考：

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Answer 2

详见下图，图片可点击放大：

Answer 3

(20)
y
=(√x+x^2.e^x +x)/x^2
=x^(-3/2)+ e^x + 1/x
y'
=-(3/2)x^(-5/2)+ e^x - 1/x^2
(21)
y=(1+x^2)arctanx/(1+x)
y'
=(1+x^2)arctanx.[1/(1+x)]' +[(1+x^2)/(1+x)].(arctanx)' +[arctanx/(1+x)].(1+x^2)'
=(1+x^2)arctanx.[-1/(1+x)^2] +[(1+x^2)/(1+x)].[1/(1+x^2)] +[arctanx/(1+x)].(2x)
=-(1+x^2)arctanx/(1+x)^2 +1/(1+x) + 2x.arctanx/(1+x)
(22)

y
=x/secx + cscx/x
=xcosx + 1/(xsinx)
y'
=x(cosx)' + cosx.(x') -[1/(xsinx)^2].(xsinx)'
=x(-sinx) + cosx.(1) -[1/(xsinx)^2].(xcosx + sinx)
=-xsinx +cosx - (xcosx + sinx)/(xsinx)^2

Answer 4

y
=(√x+x^2.e^x +x)/x^2
=x^(-3/2)+ e^x + 1/x
y'
=-(3/2)x^(-5/2)+ e^x - 1/x^2
(21)
y=(1+x^2)arctanx/(1+x)
y'
=(1+x^2)arctanx.[1/(1+x)]' +[(1+x^2)/(1+x)].(arctanx)' +[arctanx/(1+x)].(1+x^2)'
=(1+x^2)arctanx.[-1/(1+x)^2] +[(1+x^2)/(1+x)].[1/(1+x^2)] +[arctanx/(1+x)].(2x)
=-(1+x^2)arctanx/(1+x)^2 +1/(1+x) + 2x.arctanx/(1+x)
(22)

y
=x/secx + cscx/x
=xcosx + 1/(xsinx)
y'
=x(cosx)' + cosx.(x') -[1/(xsinx)^2].(xsinx)'
=x(-sinx) + cosx.(1) -[1/(xsinx)^2].(xcosx + sinx)
=-xsinx +cosx - (xcosx + sinx)/(xsinx)^2