How do you use the chain rule to differentiate #ln(tanx)#?

Question

2332 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-08T04:43:19

#(dy)/(dx)=(sec^2x)/tanx#

We can treat #f(x)=ln(tanx)# as a function in terms of #tanx#.

Let #u=tanx#, #y=lnu#

Chain rule: #(dy)/(dx)=(dy)/(du)*(du)/(dx)#

#(dy)/(du)=1/u#

#(du)/(dx)=sec^2x#

#:.(dy)/(dx)=1/(tanx)*sec^2x#

#=(sec^2x)/(tanx)#

Further simplifying...

#=(1/(cos^2x))/((sinx)/(cosx))#

#=1/(sinxcosx)#

#=2/(2sinxcosx)#

#=2/(sin2x)#

#=2csc2x#