How do you find the derivative of #(ln tan(x))^2#?

Question

Charles Anctil · Accepted Answer

How?  Use the chain rule -- twice.
And simplify the answer. (Often simplifying is the tricky part.) The derivative of some stuff squared is two times the stuff, times the derivative of that stuff.
Formally : the derivative of #[g(x)]^2#  is #2[g(x)]*g'(x)# In this case the "inside stuff", the #g(x)# is itself, the #ln# of some stuff.  So to find the derivative, we'll need the chain rule again.
The derivative of the #la# of some stuff is #1# over the stuff, times the derivative of that stuff.
Formally : the derivative of #ln[f(x)]#  is #1/f(x)*f'(x)# Finally, we need to know the derivative of #tanx#. The derivative of #(ln tan(x))^2# is   #2(ln tan(x))*1/tan(x)*sec^2(x)# Now, we've finished the calculus, but we can re-write our answer in simpler form.
#1/tan(x) = cot(x)#
and #cot(x) = cos(x)/sin(x)#. The second equality wouldn't be helpful, except that #sec(x)=1/cos(x)#. Therefore we can re-write the derivative using: #1/tan(x)*sec^2(x) = cot(x)sec^2(x)=cos(x)/sin(x)*1/cos^2(x)=1/sin(x)*1/cos(x)# Which is surely more simply written as #csc(x)sec(x)# So 
The derivative of #(ln tan(x))^2# is   #2(ln tan(x))csc(x)sec(x)# (If there's any advantage to it, we could also re-write #ln tan(x)# as #ln(sinx)-ln(cosx)#.)