Find dy/dx and d²y/dx². x = cos(2t), y = cos(t), 0 < t < π

Solution:

It is given that

x = cos(2t)

y = cos(t)

Using the chain rule, the first derivative is

dy/dx = dy/dt dt/dx = dy/dt/ dx/dt

Here

x = cos (2t)

dx/dt = -2 sin 2t = -4 sin t cos t

y = cos (t)

dy/dt = - sin (t)

dy/dx = - sin t/ -4 sin t cos t

dy/dx = sin t/ 4 sin t cos t

dy/dx = 1/4 sec t

We know that the second derivative is

d²y/dx² = dz/dx = dz/dt dt/dx = dz/dt/ dx/dt

Here z = dy/dx = 1/4 sec t

d²y/dx² = [d/dt (1/4 sec t)/ -4 sin t cos t]

d²y/dx² = (1/4 sec t tan t)/ - 4 sin t cos t

So we get

d²y/dx² = -1/16 sec³t

Therefore, dy/dx = 1/4 sect and d²y/dx² = -1/16 sec³t.