(cot^-1)(whole root (1 + cos3x)/1 - cos3x))
can someone differentiate it w. r.t x

The soluton to the problem is as follows:

d/dx (cot⁻¹(x)) = −1/(x²+1) 

d/dx (cot⁻¹ √[(1+cos(3x))/(1−cos(3x))]) 

= −1/[(1+cos(3x))/(1−cos(3x)) + 1] * d/dx √[(1+cos(3x))/(1−cos(3x))] 

= −(1−cos(3x))/2 * 1/2 [(1+cos(3x))/(1−cos(3x))]^(−1/2) * d/dx (1+cos(3x))/(1−cos(3x)) 

= −1/4 (1−cos(3x))^(3/2)/√(1+cos(3x)) * [−3sin(3x)(1−cos(3x)) − (1+cos(3x))(3sin(3x))]/(1−cos(3x))^2 

= −1/4 (1−cos(3x))^(3/2)/√(1+cos(3x)) * (−6sin(3x))/(1−cos(3x))^2 

= 3 sin(3x) / (2 √[(1−cos(3x))(1+cos(3x))]) 

= 3 sin(3x) / (2 √(1−cos²(3x))) 

= 3 sin(3x) / (2 √(sin²(3x))) 

= 3 sin(3x) / (2 |sin(3x)|)

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It is 6% because if you subtract 69-58=6 so 6% is the answer to your question.