Question

Find dy/dx for y=log (tan2x)

Answer 1

`(dy)/(dx) = (2)/(sin2x cos2x)`

Explanation:

y = log (tan 2x)

Applying chain rule:

`(dy)/(dx) = (d)/(dx) (1st function) . (d)/(dx) (2nd function). (d)/(dx)(3rd function)`

`(dy)/(dx) = (d)/(dx) log(tan 2x) . (d)/(dx) tan2x. (d)/(dx) 2x`

`(dy)/(dx) = (1)/(tan2x) * sec^(2)2x * 2`

`(dy)/(dx) = (2sec^(2)2x )/(tan2x)`

Now since
`sec^(2)x = (1)/(cos^(2)x ) \ and\ tanx = (sinx)/(cosx)`

∴
`(dy)/(dx) = (2cos2x )/(sin2x cos^22x)`

∴
`(dy)/(dx) = (2)/(sin2x cos2x)`