求高数大神帮忙做，万分感谢
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令g(x)=√sinx，则g'(x)=cosx/2√sinx
f(x)=ln[(1+g(x))/(1-g(x))]+2arctang(x)
=ln[1+g(x)]-ln[1-g(x)]+2arctang(x)
f'(x)=g'(x)/[1+g(x)]+g'(x)/[1-g(x)]+2g'(x)/[1+g^2(x)]
=2g'(x)/[1-g^2(x)]+2g'(x)[1+g^2(x)]
=4g'(x)/[1-g^4(x)]
=(2cosx/√sinx)/(1-sin^2x)
=2/cosx√sinx
f(x)=ln[(1+g(x))/(1-g(x))]+2arctang(x)
=ln[1+g(x)]-ln[1-g(x)]+2arctang(x)
f'(x)=g'(x)/[1+g(x)]+g'(x)/[1-g(x)]+2g'(x)/[1+g^2(x)]
=2g'(x)/[1-g^2(x)]+2g'(x)[1+g^2(x)]
=4g'(x)/[1-g^4(x)]
=(2cosx/√sinx)/(1-sin^2x)
=2/cosx√sinx
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