设y=ln(tanx/2)-ln1/2,则y'(π/2)=?
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![设y=ln(tanx/2)-ln1/2,则y'(π/2)=?](/uploads/image/z/10193081-41-1.jpg?t=%E8%AE%BEy%3Dln%28tanx%2F2%29-ln1%2F2%2C%E5%88%99y%27%28%CF%80%2F2%29%3D%3F)
设y=ln(tanx/2)-ln1/2,则y'(π/2)=?
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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