What is the derivative of #y=sec^2(x) + tan^2(x)#?
1 Answer
The derivative of
#4sec^2xtanx#
Process:
Since the derivative of a sum is equal to the sum of the derivatives, we can just derive
For the derivative of
#F(x) = f(g(x))#
#F'(x) = f'(g(x))g'(x)# ,
with the outer function being
#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
#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#