Question

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

Answer 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

Answer 2

To evaluate F(x) = ln(cos²(4z)), where z = π/16:

We can use the chain rule of differentiation to find the derivative of F(x) with respect to x:

F'(x) = d/dx [ln(cos²(4z))] = 1/(cos²(4z)) * d/dx [cos²(4z)] = -8 tan(4z) / (cos²(4z))

Then, we can substitute z = π/16 to find the value of F'(x) at x:

F'(x) = -8 tan(π/4) / (cos²(π/4)) = -8

Therefore, the derivative of F(x) at x is -8.

To evaluate f(t) = 4cos²(t), where t = π/4:

We can substitute t = π/4 into the expression for f(t) to get:

f(π/4) = 4cos²(π/4) = 4(1/√2)² = 2

Therefore, f(π/4) = 2.

To evaluate f(t) = 4sin⁵(2t), where t = π/8:

We can substitute t = π/8 into the expression for f(t) to get:

f(π/8) = 4sin⁵(2(π/8)) = 4sin⁵(π/4) = 4(1/√2)⁵ = 2/√2 = √2

Therefore, f(π/8) = √2.