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))

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- October 16th 2008, 04:39 PMCalcNewbDerivative Help Please
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)) - October 16th 2008, 04:47 PMChris L T521
- October 16th 2008, 04:55 PMCalcNewb
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) - October 16th 2008, 07:26 PMChris L T521
- October 16th 2008, 08:40 PMCalcNewb