## (cscx)’

### To find the derivative of csc(x), we can use the chain rule

To find the derivative of csc(x), we can use the chain rule. Recall that the chain rule states that if we have a function F(g(x)), the derivative of F(g(x)) with respect to x is given by the product of the derivative of F with respect to its argument (g'(x)) and the derivative of g with respect to x.

Let’s start with the definition of the cosecant function:

csc(x) = 1/sin(x)

To find (csc(x))’, we need to find the derivative of 1/sin(x). Let’s set y = 1/sin(x). Now, we are taking the derivative of y with respect to x.

To simplify the expression, we can rewrite y as y = sin(x)^(-1). Now, taking the derivative using the chain rule:

dy/dx = d/dx (sin(x)^(-1))

For this, we can rewrite y in exponential form:

y = sin(x)^(-1) = (sin(x))^(-1)

Let’s use u = sin(x) as the inner function. By using the power rule, we can find the derivative:

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

Now, we need to find du/dx. Since u = sin(x), the derivative is simply:

du/dx = d/dx (sin(x)) = cos(x)

Now, we have both dy/du and du/dx. We can multiply them to find dy/dx:

dy/dx = (dy/du) * (du/dx)

= (-1/sin^2(x)) * (cos(x))

= -cos(x)/sin^2(x)

Therefore, the derivative of csc(x) (or (csc(x))’) is -cos(x)/sin^2(x).

Note: It’s worth mentioning that an alternate method to find the derivative of csc(x) is by using the quotient rule directly on 1/sin(x), but the chain rule method described above is more commonly used and shows the steps in a more systematic manner.

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