Q.82. Derivatives.

Question 82.
1) If x=cos 4A+2cos 2A, y=sin 4A-2sin 2A, then
     show that dy/dx=tan A and d^2y/dx^2=-1/(8sin 3A cos^3 A).

2) if x=t-sin t and y=1-cos t, then
    show that dy/dx=cot1/2 and d^2y/dx^2=-1/(4sin^4(t/2).

Answer 82.
1)
x=cos 4A+2cos 2A, y=sin 4A-2sin 2A
=> dx/dA= -4sin4A - 4sin2A = -8 sin 3AcosA ... ( 1 )
and dy/dA = 4cos4A - 4cos2A = -8sin3A sinA ... ( 2 )

dy/dx
= (dy/dA) / (dxdA)
= [ - 8sin3A sinA ] / [ -8 sin 3AcosA ]
= tanA

d^2y/dx^2
= d/dx(dy/dx)
= d/dA(dy/dx)*dA/dx
= (sec^2A)*[ 1/[-8 sin 3AcosA ]
= - 1/[8 sin 3Acos^3A]

2)
x=t-sin t and y=1-cos t
=> dx/dt = 1 - cost and dy/dt = sint

=> dy/dx
= (dy/dt)/(dx/dt)
= sint / (1 - cost)
= 2sin(t/2)cos(t/2) / 2sin^2 (t/2)
= cot(t/2)

d^2y/dx^2
= d/dx(dy/dx)
= d/dt(dy/dx)*(dt/dx)
= -(1/2)cosec^2 (t/2) * [1/2sin^2 (t/2)]
= - 1/[4sin^4(t/2)]

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