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