why in this derivative the tan(x) doesn't distribute to the sec(x)

f(x)= sec(x)tan(x)

= sec^2(x) (sec(x)) + (sec(x)tan(x))(tan(x))

= sec^3(x) + tan^2(x) sec(x)

maybe I'm just missing an easy rule, I'm not sure though

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- Mar 3rd 2010, 06:48 PMtbenne3can someone explain to me..
why in this derivative the tan(x) doesn't distribute to the sec(x)

f(x)= sec(x)tan(x)

= sec^2(x) (sec(x)) + (sec(x)tan(x))**(tan(x))**

= sec^3(x) + tan^2(x) sec(x)

maybe I'm just missing an easy rule, I'm not sure though - Mar 3rd 2010, 07:02 PMArchie Meade
- Mar 3rd 2010, 07:33 PMtbenne3
- Mar 3rd 2010, 07:55 PMexperiment00005
You're thinking too hard. It's simply the distributive property (not) at play.

(a * b) * c = a * (b * c) = a * b * c

In this case sec x * tan x * tan x just ends up being tan^2(x) * sec(x).

Hope that helped! - Mar 3rd 2010, 08:00 PMtbenne3
- Mar 3rd 2010, 08:09 PMexperiment00005
- Mar 3rd 2010, 08:23 PMKasper
- Mar 3rd 2010, 09:03 PMtbenne3
- Mar 3rd 2010, 09:09 PMKasper
But it is!

You can also think of it like this:

Say we have , would you agree that

This is the same for , you can just think of the product as being of 3 parts, where two pieces are the same.

We can write it like this too if it helps;

EDIT : Just hit me you might be talking about the tan not distributing to the term on the right. And that is just due to the nature of the product rule.

Just focus on , where for this problem; and .

Just plug away and take derivatives where appropriate. - Mar 4th 2010, 01:22 AMArchie Meade
For

Using the product rule of differentiation,

as

and

This is

or

which has sec(x) as a factor, so it may also be written

When you asked, why is

not equal to

it's because you've brought in an extra tan(x) that does not belong