Question

derivative of ln(tanx)

Answer 1

how do you get 2csc2x?

Answer 2

Using chain rule:
d[ln(tanx)]/dx=1/tanx * d(tanx)/dx

Answer 3

so you get cotxsec^2(x), but how do you convert it to 2csc2x?

Answer 4

cotx=1/tanx=cosx/sinx
sec^x=1/cos^x
f'(x)=1/(sinx*cosx)
remember 2*sinx*cosx=sin2x?

Answer 5

Where do yo get the two?

Answer 6

the two in front of the csc2x

Answer 7

Hmm, this is trigonometric identity. We can prove it, but here's a simple way to see it IS correct:
sin(2A)=2*sinA*cosA
let's say A=30 deg
sin(2*30)=sin60=sqrt(3)/2
2*sin30*cos30=2*1/2*sqrt(3)/2=sqrt(3)/2

Answer 8

sorry, I mean where does the two in front of "2csc2x" come from? I get how the csc2x works, just the constant in front.