What is the derivative of #3sec^2x + tan^2x#?

1 Answer
Jan 7, 2016

#8sec^2(x)tan(x)#

Explanation:

Use partial differentiation.
Looking at the first term, let #sec(x) = t#
Then #d/dt 3t^2 = 6t#
#d/dx sec(x) = sec(x)tan(x)#
#d/dt*d/dx = 6sec(x)sec(x)tan(x) = 6sec^2(x)tan(x)#

for the second term, let #tan(x) = t#
#d/dt t^2 = 2t#
#d/dx tan(x) = sec^2(x)#
#d/dt*d/dx = 2tan(x)sec^2(x)#

Then the whole derivative is
#6sec^2(x)tan(x) + 2sec^2(x)tan(x)#
#=8sec^2(x)tan(x)#