How derivative of (1/cot(x)) = sec^2(x) ?

2 Answers
May 10, 2018

#d/dx 1/cotx = d/dxtgx = sec^2x# because, #tgx = sinx/cosx#, like this you have to derive using the rule of the quotient. See bellow:

Explanation:

#d/dx tgx = d/dx sinx/cosx = (sinx' cosx - sinx cosx')/(cosx)^2 = (cosxcosx - sinx(-sinx))/(cosx)^2 = (cos^2x + sin^2x)/(cosx)^2 = 1/ cos^2x = sec^2x#.

I hope to helped you!

May 10, 2018

the answer
#tan^2x+1=2sec^2x*tanx#

Explanation:

show below

#d/dx[1/cotx]=d/dx[sec^2x]#

#cotx=cosx/sinx#

#d/dx[sinx/cosx]=d/dx[sec^2x]#

#[cosx*(cosx)-sinx*(-sinx)]/cos^2x=2secx*secx*tanx#

#[cos^2x+sin^2x]/cos^2x=2sec^2x*tanx#

#(sin^2x/cos^2x)+(cos^2x/cos^2x)=2sec^2x*tanx#

#tan^2x+1=2sec^2x*tanx#