Question

Find the derivative of


`\\ \rm\Rrightarrow y=(sinx+cosx)/(sinx-cosx)`

Proper explanation is mandatory


Note:-
Spams/sHort/wrong answers will be deleted on the spot .

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Answer 1

⇒
`y'=-(2)/((six-cosx)^(2) )`

Explanation:

⇒Answer in the attachment.

Answer 2

y' = 1 / sinxcosx

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]