Answer
d/dx sec(x^2)= sec(x^2)tan(x^2)2xddxsec(x2)=sec(x2)tan(x2)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)sec(x2), you can quite easily identitfy the "inner" and "outer" functions present to use the chain rule. The "inner" function is x^2x2, because it is composed/inside of in the secsec.
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
