How do you differentiate #f(x)=sec^2x-tan^2x#?

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1 Answer
F. Javier B.
Apr 18, 2018

See below

Explanation:

#(df)/dx=2secx(dsecx)/dx-2tanx(dtanx)/dx=2secxtanxsecx-2tanxsec^2x=2sec^2xtanx-2sec^2xtanx=0#

This obvious result is the consecuence of the fact

#sec^2x-tan^2x=1/(cos^x)-sin^2x/cos^2x=(1-sin^2x)/cos^2x=cos^2x/cos^2x=1#

Therefore #f(x)# is a constant function for every #x#. For this reason his derivative is zero

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