`y = (1 - sec(x))/tan(x)` Differentiate.

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Chapter 3, 3.3 - Problem 14 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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`y = (1 - sec(x))/tan(x)`

`=(1 /tan(x))- (sec(x)/tan(x))`

`y= cot(x) - csc(x)`

so now the y' is as follows

`y' = d/(dx) (cot(x) - csc(x))`

`= d/(dx) (cot(x)) - d/(dx)(csc(x))`

as

`d/(dx) (cot(x))`

`=d/(dx) (cos(x)/sin(x))`

`= ( ((cosx *(d/(dx) sin x)) -((d/(dx) cos x * sinx)))/(sin^2 x)`

`= (-cos^2 x -sin^2 x)/(sin^2 x)`

`= - 1/(sin^2 x)`

`= -csc^2(x)`

and

`d/(dx) (csc(x))`

`=d/(dx) (1/sin(x))`

`= ((d/(dx) sin x)*1 - sinx * d/dx(1))/(sin^2(x))`

`= -cosx/sin^2(x)`

`=-csc x cot(x)`

so,

`y' =d/(dx) (cot(x)) -d/(dx)(csc(x))`

`=-csc^2(x)-csc x cot(x)`

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