differentiate sinx+cosx/sinx-cosx w.r.t.x

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You need to differentiate using the quotient law such that:

`((sinx+cosx)/(sinx-cosx))'= ((sinx+cosx)'*(sinx-cosx) - (sinx+cosx)*(sinx-cosx)')/((sinx-cosx)^2)`

`((sinx+cosx)/(sinx-cosx))'= ((cos x - sin x)*(sinx-cosx) - (sinx + cosx)*(cos x + sin x))/((sinx-cosx)^2)`

`((sinx+cosx)/(sinx-cosx))'= (-(sin x - cos x)^2 - (sinx + cosx)^2)/((sinx-cosx)^2)`

`((sinx+cosx)/(sinx-cosx))'= -(sin^2 x - 2sin x*cos x + cos^2 x + sin^2 x + 2sin x*cos x + cos^2 x)/((sinx-cosx)^2)`

Reducing like terms yields :

`((sinx+cosx)/(sinx-cosx))' = -(sin^2 x + cos^2 x + sin^2 x + cos^2 x)/((sinx-cosx)^2)`

You need to remember that `sin^2 x + cos^2 x = 1`  such that:

`((sinx+cosx)/(sinx-cosx))' = (-2)/((sinx-cosx)^2)`

Hence, differentiating with respect to x yields `((sinx+cosx)/(sinx-cosx))' = (-2)/((sinx-cosx)^2).`

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