What is derivative of cosec(ax+b) ? please explain it step by step

(csc(ax+b))'=-a csc(ax+b)cot(ax+b)

Step-by-step explanation:

We are asked to find the derivative of csc(ax+b). I assume you are asked to find the derivative with respect to x. If the assumption is wrong, please note so in the comments.

To do this we must use chain rule.

This will require us to know the derivative (w.r.t. x) of inside function, ax+b, and outside function, csc(x).

(ax+b)'=(ax)'+(b)'

(ax+b)'=a(x)'+0

(ax+b)'=a(1)+0

(ax+b)'=a

(csc(x))'=-csc(x)cot(x)

The chain rule states:

(f(g(x)))'=g'(x)×f'(g(x))

So our conclusion is:

(csc(ax+b))'=a×-csc(ax+b)cot(ax+b)

(csc(ax+b))'=-a csc(ax+b)cot(ax+b)

choose (5, 1.5) and (-1.5,-3.5)

step-by-step explanation:

5 - 1.5 =3.5

-1.5 - (-3.5) = 2

3.5 + 2 =5.5

hope this !