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)]]