This is the problem I have f(x)= x^3 ( sec(x) )
And this is what I think the answer is. Is it correct? If so could I possibly just distribute the sec(x) through (sec(x) tan(x) )? Thanks ahead of time.
f'(x)= 6x^2 (sec(x)) (sec(x) tan(x))
This is the problem I have f(x)= x^3 ( sec(x) )
And this is what I think the answer is. Is it correct? If so could I possibly just distribute the sec(x) through (sec(x) tan(x) )? Thanks ahead of time.
f'(x)= 6x^2 (sec(x)) (sec(x) tan(x))
ok yes I do understand how to use the product rule. I used the product rule but got something totally different. Let me try to show you my steps.
"first times derivative of second + second times derivative of the first" this is what our calc teacher told us to remember for product rule. so here it goes
D=derivative
x^3 D(sec(x)) + sec(x) D(x^3)
so I get
X^3 (sec(x) tan(x)) + sec(x) (3x^2)
this is almost what you have I just dont see how you got the other sec^2(x)