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)

good luck!

cutie~

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.