What is the derivative of #y=sec^2(x) + tan^2(x)#?

The derivative of #y=sec^2x + tan^2x# is:

#4sec^2xtanx#

Process:

Since the derivative of a sum is equal to the sum of the derivatives, we can just derive #sec^2x# and #tan^2x# separately and add them together.

For the derivative of #sec^2x#, we must apply the Chain Rule:

#F(x) = f(g(x))#
#F'(x) = f'(g(x))g'(x)#,

with the outer function being #x^2#, and the inner function being #secx#. Now we find the derivative of the outer function while keeping the inner function the same, then multiply it by the derivative of the inner function. This gives us:

#f(x) = x^2#
#f'(x) = 2x#

#g(x) = secx#
#g'(x) = secxtanx#

Plugging these into our Chain Rule formula, we have:

#F'(x) = f'(g(x))g'(x)#,
#F'(x) = 2(secx)secxtanx = 2sec^2xtanx#

Now we follow the same process for the #tan^2x# term, replacing #secx# with #tanx#, ending up with:

#f(x) = x^2#
#f'(x) = 2x#

#g(x) = tanx#
#g'(x) = sec^2x#

#F'(x) = f'(g(x))g'(x)#,
#F'(x) = 2(tanx)sec^2x = 2sec^2xtanx#

Adding these terms together, we have our final answer:

#2sec^2xtanx + 2sec^2xtanx#

= #4sec^2xtanx#