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олл26
@олл26
July 2022
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f(z)＝ln sin z ;f'(π/6)
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армения20171
F'(lnsinz)= 1/sinz*sin'z=cosz/sinz
f'(π/6)=(cosπ/6)/sinπ/6=√3/2:1/2=√3/2*2=√3
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Answers & Comments
f'(π/6)=(cosπ/6)/sinπ/6=√3/2:1/2=√3/2*2=√3