设y=ln(tanx/2)-ln1/2,则y'(π/2)=?

设y=ln(tanx/2)-ln1/2,则y'(π/2)=?

y'=1/tanx/2*(tanx/2)'
=1/tanx/2*sec²x/2*(x/2)'
=cosx/2/sinx/2*1/cos²x/2*1/2
=1/(2sinx/2cosx/2)
=1/sinx
所以原式=1/sinπ/2=1

y'=(1/2*sec^2(x/2))/tan(x/2)
y'(π/2)=1

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