How do you differentiate the function #y=tan[ln(ax + b)]#?

1 Answer
Feb 17, 2015

Here you have a function (#tan#) of a function (#ln#) of a function (#ax+b#).
You can use the Chain Rule where you derive each function leaving the "nested" one as it is and multiply the derivative together.
So you get:
#tan(x)# derived gives you: #1/cos^2(x)#
#ln(x)# derived gives you: #1/x#
#ax+b# derived gives you: #a#.

So finally:
#y'=1/(cos^2(ln(ax+b)))*1/(ax+b)*a#

This can also be written as

#y'=(asec^2(ln(ax+b)))/(ax+b)#