Trig question If g(x)=cscx+cotx, then g'(pi/6)=

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You need to find derivative of the function g(x) such that:

`g'(x) = (csc x + cot x)'`

You need to remember that cosecant function is the reverse of sine function and cotangent function is a rational function cot x = cos x/sin x.

`g'(x) = (1/(sin x) + cos x/sin x)'`

`g'(x) = ((1 +cos x)/(sin x))'`

`g'(x) = ((1 +cos x)'*(sin x) - (1 +cos x)*(sin x)')/(sin^2 x)`

`g'(x) = ((-sin x)*(sin x) - (1 +cos x)*(cos x))/(sin^2 x)`

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

You need to remember that `sin^2 x + cos ^2 x = 1` , hence `-sin^2 x- cos ^2 x = -1` => `g'(x) = -(1 + cos x)/(sin^2 x)`

You need to substitute `pi/6 ` for x in equation of derivative such that:

`g'(pi/6) = -(1 + cos (pi/6))/(sin^2 (pi/6))`

`g'(pi/6) = -(1 + sqrt3/2)/(1/4)`

`g'(pi/6) = -4(1 + sqrt3/2)`

`g'(pi/6) = -4 - 2sqrt3`

Hence, evaluating derivative of g(x) at `x = pi/6`  yields `g'(pi/6) = -4 - 2sqrt3` .

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