find d^2y of y=4sec 2cscx
To find the second derivative of y with respect to x, let's start by finding the first derivative.
First, let's express y = 4sec(2csc(x)) in terms of sin and cos.
Recall that sec(x) = 1 / cos(x) and csc(x) = 1 / sin(x).
So, y = 4sec(2csc(x)) can be written as y = 4 / cos(2 / sin(x)).
Now, we can find the first derivative of y with respect to x.
dy/dx = d(4 / cos(2 / sin(x))) / dx
To find this derivative, we can use the chain rule.
The chain rule states that if we have a composite function u(v(x)), then the derivative of u with respect to x is given by du/dx = du/dv * dv/dx.
Let's apply the chain rule step by step.
Let u = 4, and v = 2 / sin(x).
Then, we have y(u(v(x))) = u / cos(v) = 4 / cos(2 / sin(x)).
Now, let's find the derivative of u with respect to v.
du/dv = 0. (Derivative of a constant is 0.)
Next, let's find the derivative of v with respect to x.
dv/dx = -2 * (csc(x))^2 * cot(x). (Derivative of 2 / sin(x) using trigonometric identities.)
Now, using the chain rule, we can find dy/dx.
dy/dx = du/dv * dv/dx
= 0 * (-2 * (csc(x))^2 * cot(x))
= 0. (The first derivative turns out to be 0.)
Since the first derivative is 0, it means that the function y is a constant.
Now, to find the second derivative, we need to differentiate dy/dx again.
d^2y/dx^2 = d(0) / dx
= 0. (Derivative of a constant is 0.)
Therefore, the second derivative of y with respect to x is 0.