Question

Please help me with finding the derivative of the following problem...

y=ln(tanx)

Answer 1

\[\frac{ dy }{ dx }=\frac{ 1 }{ tanx}\times \sec x\]

Answer 2

oops shouldbe sec squared x

Answer 3

sec(x)^2

Answer 4

\[=\frac{ \sec ^{2}x }{ tanx }\]

Answer 5

answer in the book is 2csc2x

Answer 6

sec squared x is the same as 1 over cos squared x and 1 over tan x is cos x over sin x so you end up with 1 over cos x sin x which is the same as a half sin 2x and 1 over a half sin 2x is 2 csc 2x

Answer 7

sorry in the middle there i just wanted to say cos x sin x is a half sin 2x