Differentiate with respect to X \sqrt{ (cos2x)/(1 +sin2x ) } ​

Question

Differentiate with respect to X

`\sqrt{ (cos2x)/(1 +sin2x ) }`
​

Answer 1

Power and chain rule (where the power rule kicks in because
`\sqrt x=x^(1/2)`):

`\left(\sqrt{(\cos(2x))/(1+\sin(2x))}\right)'=\frac1{2\sqrt{(\cos(2x))/(1+\sin(2x))}}\left((\cos(2x))/(1+\sin(2x))\right)'`

Simplify the leading term as

`\frac1{2\sqrt{(\cos(2x))/(1+\sin(2x))}}=(√(1+\sin(2x)))/(2√(\cos(2x)))`

Quotient rule:

`\left((\cos(2x))/(1+\sin(2x))\right)'=((1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))')/((1+\sin(2x))^2)`

Chain rule:

`(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)`

`(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)`

Put everything together and simplify:

`(√(1+\sin(2x)))/(2√(\cos(2x)))((1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x)))/((1+\sin(2x))^2)`

`=(√(1+\sin(2x)))/(2√(\cos(2x)))(-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x))/((1+\sin(2x))^2)`

`=(√(1+\sin(2x)))/(2√(\cos(2x)))(-2\sin(2x)-2)/((1+\sin(2x))^2)`

`=-(√(1+\sin(2x)))/(√(\cos(2x)))(\sin(2x)+1)/((1+\sin(2x))^2)`

`=-(√(1+\sin(2x)))/(√(\cos(2x)))\frac1{1+\sin(2x)}`

`=-\frac1{√(\cos(2x))}\frac1{√(1+\sin(2x))}`

`=\boxed{-\frac1{√(\cos(2x)(1+\sin(2x)))}}`