Question

differentiation of root under 1-cos2x

Answer 1

ddxcos1−cos2x1+cos2x−−−−−−−−√ddxcos1−cos2x1+cos2x


Remember: you have to multiply the derivative of the outside by the derivative of the inside 
= −sin1−cos2x1+cos2x−−−−−−−−√(ddx1−cos2x1+cos2x−−−−−−−−√)−sin1−cos2x1+cos2x(ddx1−cos2x1+cos2x)


Now, let's find the derivative of the inside
ddx1−cos2x1+cos2x−−−−−−−−√ddx1−cos2x1+cos2x = 121−cos2x1+cos2x−−−−−−−−√(ddx1−cos2x1+cos2x)121−cos2x1+cos2x(ddx1−cos2x1+cos2x)


And find the derivative of the inside one more time and simplify
ddx1−cos2x1+cos2xddx1−cos2x1+cos2x = 2sin2x1+cos2x+2(1−cos2x)(sin2x)(cos(2x)+1)2=4sin2x(cos(2x)+1)22sin2x1+cos2x+2(1−cos2x)(sin2x)(cos(2x)+1)2=4sin2x(cos(2x)+1)2


Plug it back into the original 
ddxcos1−cos2x1+cos2x−−−−−−−−√ddxcos1−cos2x1+cos2x


= −sin1−cos2x1+cos2x−−−−−−−−√x121−cos2x1+cos2x−−−−−−−−√x4sin2x(cos(2x)+1)2−sin1−cos2x1+cos2xx121−cos2x1+cos2xx4sin2x(cos(2x)+1)2


simplify
= −2sin(2x)sin−cos(2x)−1cos(2x)+1−−−−−−−−−−−√(cos(2x)+1)2×−cos(2x)−1cos(2x)+1

Answer 2

Answer:

Step-by-step explanation:

=(3+2sin2x)(sin2x)2-(1-cos2x)(2sin2x).2 /(3+2sin2x)2 =2(2+3sin2x-2cos2x)/(3+2sin2x)2