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
It looks like you have

$\displaystyle f(x) = sec(x)tan(x)$

$\displaystyle f'(x) = sec^2(x) \cdot [sec(x) + sec(x)tan(x)] \cdot tan(x)$

This is not the derivative! Gotta watch the distributions, Product Rule : $\displaystyle \frac{d}{dx}(f(x) \cdot g(x)) = f'(x)g(x) + g'(x)f(x)$

$\displaystyle f(x) = sec(x)tan(x)$

$\displaystyle f'(x) = [sec(x)]' \cdot tan(x) + [tan(x)]' \cdot sec(x)$

$\displaystyle f'(x) = [sec(x)tan(x)] \cdot tan(x) + [sec^2(x)] \cdot sec(x)$

$\displaystyle f'(x) = sec(x)tan^2(x) + sec^3(x)$

Does this help? - 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 $\displaystyle A B B$, would you agree that

$\displaystyle A B B = A B^2$

This is the same for $\displaystyle tan(x)$, 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;

$\displaystyle sec(x) tan(x) tan(x) = sec(x) [tan(x)]^2 = sec(x) tan^2(x)$

EDIT : Just hit me you might be talking about the tan not distributing to the $\displaystyle sec^3x$ term on the right. And that is just due to the nature of the product rule.

Just focus on $\displaystyle \frac{d}{dx}(f(x) \cdot g(x)) = f'(x)g(x) + g'(x)f(x)$, where for this problem; $\displaystyle f(x) = sec(x)$ and $\displaystyle g(x) = tan(x)$.

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

$\displaystyle f(x)=sec(x)tan(x)$

Using the product rule of differentiation,

$\displaystyle f'(x)=sec(x)\frac{d}{dx}tan(x)+tan(x)\frac{d}{dx}s ec(x)$

$\displaystyle =sec(x)\left[sec^2(x)\right]+tan(x)\left[sec(x)tan(x)\right]$

as

$\displaystyle \frac{d}{dx}tan(x)=sec^2(x)$

and

$\displaystyle \frac{d}{dx}sec(x)=sec(x)tan(x)$

This is

$\displaystyle sec^3(x)+sec(x)tan^2(x)$

or

$\displaystyle sec(x)sec^2(x)+sec(x)tan^2(x)$

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

$\displaystyle sec(x)\left(sec^2(x)+tan^2(x)\right)$

When you asked, why is

$\displaystyle sec(x)tan(x)tan(x)$

not equal to

$\displaystyle tan(x)sec(x)tan^2(x)$

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

$\displaystyle tan(x)sec(x)tan^2(x)=tan(x)sec(x)tan(x)tan(x)$