Differentiate root 1+tanx/1-tanx?

1 Answer
Jun 8, 2018

#dy/dx= sec^2(pi/4+x)/[2sqrt[tan(pi/4+x)]]#

Explanation:

Let
#y=sqrt[(1+tanx)/(1-tanx)]#

#y=sqrt[(1+tanx)/(1-(1xxtanx))]#

#y=sqrt[(tan(pi/4)+tan(x))/(1-tan(pi/4)tan(x))]#

#y=sqrt[tan(pi/4+x)]#

Now we can find out the #dy/dx#.
We have to apply chain rule to calculate derivative.

#dy/dx= 1/[2sqrt[tan(pi/4+x)]]xxd(.tan(pi/4+x))#

#dy/dx= sec^2(pi/4+x)/[2sqrt[tan(pi/4+x)]]xxd(pi/4+x)#

#dy/dx= sec^2(pi/4+x)/[2sqrt[tan(pi/4+x)]]#