Question

Find the derivative of

Answer 1

Answer:

⇒$y'=-\frac{2}{(six-cosx)^{2} }$

Step-by-step explanation:

⇒Answer in the attachment.

Answer 2

Answer:

y' = 1 / sinxcosx

Step-by-step explanation:

Given :

y = sinx + cosx / sinx - cosx

Applying quotient rule :

y' = [(sinx - cosx)(cosx - sinx) - (sinx + cosx)(cosx + sinx)] / (sinx -

cosx)²

y' = [sinxcosx - cos²x - sin²x + sinxcosx - (sinxcosx + cos²x + sin²x + sinxcosx)] / (sinx - cosx)²

y' = 2sinxcosx - 2sinxcosx - 2cos²x - 2sin²x / (sinx -cosx)²

y' = -2(sin²x + cos²x) / (sinx - cosx)²

y' = -2 / (sinx - cosx)²

y' = -2 / sin²x - 2sinxcosx + cos²x

y' = -2 / -2sinxcosx

y' = 1 / sinxcosx  [⇒ Final answer]