Differentiate
Y = 1/SinX^2
the way you typed it, Wolfram read it as
http://www.wolframalpha.com/input/?i=+Y+%3D+1%2FSinX%5E2
and y = (sinx)^-2
dy/dx = -2(sinx)^-3 (cosx)
= -2cotx /sin^2 x
or
= -2cotx csc^2 x
if you meant:
y = 1/ sin(x^2)
then y = sin(x^2)^-1
dy/dx = -1sin(x^2)^-2 cos(x^2) (2x)
= -2x cot(x^2) csx(x^2)
To differentiate the function Y = 1/sin(x^2), we can use the quotient rule of differentiation. The quotient rule states that if we have a function in the form f(x) = g(x) / h(x), the derivative of f(x) can be found using the formula:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
Now let's apply the quotient rule to differentiate Y = 1/sin(x^2):
First, we identify that g(x) = 1 and h(x) = sin(x^2).
Next, we find the derivatives of g(x) and h(x):
g'(x) = 0 (the derivative of a constant is zero)
To find h'(x), we need to apply the chain rule.
Let u = x^2, so h(x) = sin(u).
Using the chain rule, h'(x) = cos(u) * d/dx(u) = cos(x^2) * 2x.
Now we have all the necessary components to apply the quotient rule:
Y' = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
= (0 * sin(x^2) - 1 * cos(x^2) * 2x) / (sin(x^2))^2
= -2x cos(x^2) / sin^2(x^2)
Therefore, the derivative of Y = 1/sin(x^2) is -2x cos(x^2) / sin^2(x^2).