математика

F(x)=ln cos^2*4z, z=pi/16 f(t)=4cos^2*t, t=pi/4 f(t)=4sin^5*2t, t=pi/8

Оставить ответ
1

Ответ №1

f`(z)=-8cos4zsin4z/cos²4z=-4sin8z/cos²4z
f`(π/16)=-4sinπ/2/cos²π/4=-4:1/2=-8

f`(t)=-8costsint=-4sin2t
f`(π/4)=-4sinπ/2=-4

f`(t)=40sin^4(2t)cos(2t)
f`(π/8)=40sin^4(π/4)cos(π/4)=40*1/4*√2/2=5√2

f`(z)=cosz*e^sinz -sinz*e^cosz
f`(π/2)=cosπ/2*e^sinπ/2 -sinπ/2*e^cosπ/2=0-1=-1

f`(y)=3/tg3ycos²3y=3cos3e/sin3ycos²3y=3/sin3ycos3y=6/sin6y
f`(π/12)=6/sinπ/2=6/1=6

Знаете ответ?