Answer :
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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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)|)
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!