Given that x=cos^2t and y=ln(sint) find d^2y/dx^2 at the point t=π/4

y = ln(sint)

dy/dt = (1/sint)(cost) = cost/sint

for x= cos^2 t = (cost)^2
dx/dt = 2cost(-sint)

(dy/dt) / (dx/dt) = (cost/sint) / (-2sintcost)
dy/dx = -1/(2 sin^2 t) = (-1/2)(sint)^-2

d(dy/dx) / dx = (-2)(-1/2)(sint)^-3 (cost)
= cost / sin^3 t
sub in t = π/4

d(dy/dx) / dx = (√2/2) / (√2/2)^3
= 1/(√2/2)^2
= 1/(2/4) = 2

x=cos^2t

y=ln(sint)

dy/dt = cost/sint
dx/dt = -2cost sint

dy/dx = dy/dt / dx/dt

= (cost/sint)/(-2cost sint)
= -1/2 csc^2(t)

d^2y/dx^2 = d/dx(dy/dx)
= d/dt (dy/dx) / dx/dt
= (-csct * csct cott)/(-2cost sint)
= 1/2 csc^2(t) cot(t) * sect csct
= 1/2 csc^3(t)

To find the second derivative of y with respect to x, we need to use the chain rule. The chain rule states that if y is a function of u, and u is a function of x, then dy/dx = (dy/du) * (du/dx).

First, let's find the first derivative of y with respect to t (dy/dt):
dy/dt = d/dt (ln(sin(t)))

To find d/dt (ln(sin(t))), we use the chain rule again. Let u = sin(t).
Then, du/dt = cos(t).

Using the chain rule:
dy/dt = (1/u) * du/dt
dy/dt = (1/sin(t)) * cos(t)
dy/dt = cot(t)

Next, we need to find dx/dt:
x = cos^2(t)
Taking the derivative of both sides with respect to t, we get:
dx/dt = d/dt (cos^2(t))

To find d/dt (cos^2(t)), we use the chain rule. Let u = cos(t).
Then, du/dt = -sin(t).

Using the chain rule:
dx/dt = 2(cos(t)) * du/dt
dx/dt = 2cos(t) * (-sin(t))
dx/dt = -2sin(t)cos(t)

Now we can find the derivative of y with respect to x (dy/dx) using the chain rule:
dy/dx = (dy/dt) / (dx/dt)
dy/dx = cot(t) / (-2sin(t)cos(t))
dy/dx = -(1/2) * cot(t)csc(t)

Finally, we can find the second derivative of y with respect to x (d^2y/dx^2):
d(dy/dx) / dt = d/dt (-(1/2) * cot(t)csc(t))
Using the quotient rule, we have:
d^2y/dx^2 = [d/dt (-(1/2))] * cot(t)csc(t) + (1/2) * [d/dt (cot(t)csc(t))]

Taking the derivatives, we get:
[d/dt (-(1/2))] * cot(t)csc(t) = 0 (since the derivative of a constant is zero)
(1/2) * [d/dt (cot(t)csc(t))] = (1/2) * (-csc^2(t) - cot^2(t))

Now, substitute t = π/4 into the second derivative expression to find the value at that point:
d^2y/dx^2 = (1/2) * (-csc^2(π/4) - cot^2(π/4))

By using the trigonometric identity csc(π/4) = √2 and cot(π/4) = 1, we can simplify further:
d^2y/dx^2 = (1/2) * (-(1/2) - 1)
d^2y/dx^2 = (1/2) * (-3/2)
d^2y/dx^2 = -3/4

Therefore, at t = π/4, the second derivative of y with respect to x is -3/4.