Find dydx if y=csc(2x−−√) .

Answer 1

To find the derivative of y with respect to x, we can use the chain rule. Let's denote u = 2x - √(x), then y = csc(u).

The derivative of y with respect to x, dy/dx, can be found as:

dy/dx = dy/du * du/dx

We know that du/dx = 2 - 1/(2√(x)).

To find dy/du, we can rewrite y = csc(u) as y = (1/sin(u)), and then find its derivative with respect to u:

(dy/du) = d(1/sin(u)) / du

Applying the chain rule, we have:

(dy/du) = -1/(sin^2(u)) * (du/du)

(dy/du) = -1/(sin^2(u))

Now, we can substitute du/dx and dy/du back into the chain rule equation:

dy/dx = dy/du * du/dx
= (-1/(sin^2(u))) * (2 - 1/(2√(x)))

Substituting back u = 2x - √(x):

dy/dx = (-1/(sin^2(2x - √(x)))) * (2 - 1/(2√(x)))

Therefore, dy/dx = (-1/(sin^2(2x - √(x)))) * (2 - 1/(2√(x)))