Question

How to find the derivative of (cosec^2x) ?

Answer 1

`d/dx (csc^2x) = -2cotxcsc^2x`

Explanation:

As:

`csc^2x = 1/sin^2x`

using the chain rule:

`d/dx (csc^2x) = -2/sin^3x d/dx sinx = -(2cosx)/sin^3x = -2cotxcsc^2x`

Answer 2

`-\frac{2cos(x)}{sin^3(x)}`

Explanation:

I'm assuming you mean `csc^2(x)` (please format your questions!!)

If so, you have by definition

`csc(x) = 1/sin(x) \implies csc^2(x) = 1/sin^2(x)`

At this point, you can apply the rule

`d/dx \frac{1}{f(x)} = -\frac{f'(x)}{f^2(x)}`

(this is a simplified version of the generic quotient rule, since the denominator is `1` and vanishes when derived)

So, you have

`-\frac{f'(x)}{f^2(x)} = -\frac{2cancel(sin(x))cos(x)}{sin^{cancel(4)3}(x)} = -\frac{2cos(x)}{sin^3(x)}`

Answer 3

`-2cotxcsc^2x`

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`"given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"`

`"here "y=csc^2x=1/sin^2x=(sinx)^-2`

`rArrdy/dx=-2(sinx)^-3xxd/dx(sinx)`

`color(white)(rArrdy/dx)=-(2cosx)/sin^3x`

`color(white)(rArrdy/dx)=-2xxcosx/sinx xx1/sin^2x`

`color(white)(rArrdy/dx)=-2cotxcsc^2x`