# can someone explain to me..

• Mar 3rd 2010, 06:48 PM
tbenne3
can 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 PM
Quote:

Originally Posted by tbenne3
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

Do you mean, why is it not

$\displaystyle Sec(x)\left(Sec^2(x)+Tan^2(x)\right)$

Yes, that's an equivalent expression
• Mar 3rd 2010, 07:33 PM
tbenne3
Quote:

Originally Posted by Archie Meade
Do you mean, why is it not

$\displaystyle Sec(x)\left(Sec^2(x)+Tan^2(x)\right)$

Yes, that's an equivalent expression

I was wondering why it wouldn't be

(sec(x)tan(x))(tan(x))

=
tan(x)sec(x)tan^2(x)
• Mar 3rd 2010, 07:55 PM
experiment00005
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 PM
tbenne3
Quote:

Originally Posted by experiment00005
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!

Ok but why is the sec(x) not multiplied by the tan(x).. It looks as if the tan(x) is only distributing to the tan(x) within the parenthesis instead of both sec(x) and tan(x)
• Mar 3rd 2010, 08:09 PM
experiment00005
Quote:

Originally Posted by tbenne3
Ok but why is the sec(x) not multiplied by the tan(x).. It looks as if the tan(x) is only distributing to the tan(x) within the parenthesis instead of both sec(x) and tan(x)

You could plug in some values to prove that it doesn't work. Try pi/6.

( sec(pi/6) * tan(pi/6) ) * tan (pi/6) = (2/3) *(sqrt(3)/3) = 2 sqrt(3) / 9

which does not equal (sec(pi/6) *(tan(pi/6) ) * (tan^2(pi/6)) = (2/3)*(1/3)
= 2/9
• Mar 3rd 2010, 08:23 PM
Kasper
Quote:

Originally Posted by tbenne3
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?
• Mar 3rd 2010, 09:03 PM
tbenne3
Quote:

Originally Posted by Kasper
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?

Yes but why in the third step is the tan(x) only distributed to the tan(x) giving you tan^2(x). Why is it not distributed to the sec(x) also?
• Mar 3rd 2010, 09:09 PM
Kasper
Quote:

Originally Posted by tbenne3
Yes but why in the third step is the tan(x) only distributed to the tan(x) giving you tan^2(x). Why is it not distributed to the sec(x) also?

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 AM
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

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