What is the derivative of #y=tan(x)# ?

1 Answer
Aug 18, 2014

The derivative of #tanx# is #sec^2x#.

To see why, you'll need to know a few results. First, you need to know that the derivative of #sinx# is #cosx#. Here's a proof of that result from first principles:

Once you know this, it also implies that the derivative of #cosx# is #-sinx# (which you'll also need later). You need to know one more thing, which is the Quotient Rule for differentiation:

Once all those pieces are in place, the differentiation goes as follows:

#d/dx tanx#
#=d/dx sinx/cosx#

#=(cosx . cosx-sinx.(-sinx))/(cos^2x)# (using Quotient Rule)

#=(cos^2x+sin^2x)/(cos^2x)#

#=1/(cos^2x)# (using the Pythagorean Identity)

#=sec^2x#