What is the derivative of #sec(x^2)#?

Question

Charles Arocha · Accepted Answer

Answer
#d/dx sec(x^2)= sec(x^2)tan(x^2)2x# Explanation To solve this question, you would need to use the chain rule (and later on, the quotient rule as well).  In #sec (x^2)#, you can quite easily identitfy the "inner" and "outer" functions present to use the chain rule. The "inner" function is #x^2#, because it is composed/inside of in the #sec#. The chain rule is:
#F'(x)=f'(g(x))(g'(x))# Or, in words: the derivative of the outer function (with the inside function left alone!) times the derivative of the inner function. To make things simpler, I'm going to denote #u=x^2# for now because at one point, we will have to leave the inner function alone, but you don't have to do this. If you do, make sure you sub #u=x^2# back in at the end so that all your variables are in terms of #x#. Steps
1) The derivative of the outer function (with the inside function #u# left alone!)
#d/dx sec (u)= d/dx 1/cos(u)#
NB:  Unless you have the derivative of #sec# memorized, you will have to do a quotient rule here to find its derivative. We know that #sec(u)=1/cos(u)# from Trig. So...
#d/dx sec (u)
= d/dx 1/cos(u)
= (cos (u)(0) - 1 (-sin (u)))/( cos (u)cos (u))= sin (u)/ (cos (u) cos (u))
= sin(u)/cos(u) *(1/cos( u))
= tan (u) sec(u)=sec (u) tan (u)# 2) The derivative of the inner function:
#d/dx x^2=2x# 3) Combining the two steps to give the actual derivative: #d/dx sec(x^2)= sec(x^2)tan(x^2)2x#