Differentiate root 1+tanx/1-tanx?

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Answer 1 · 2018-06-08T05:12:55

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

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)]]#