can someone explain to me..

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March 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
March 3rd 2010, 07:02 PM
Archie Meade

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

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

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

Quote:

Originally Posted by Archie Meade

Do you mean, why is it not

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

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)
March 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!
March 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)
March 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
March 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

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

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

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

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

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

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

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

Does this help?
March 3rd 2010, 09:03 PM
tbenne3

Quote:

Originally Posted by Kasper

It looks like you have

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

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

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

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

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

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

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

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?
March 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 $A B B$ , would you agree that

$A B B = A B^2$

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

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

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

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

Just plug away and take derivatives where appropriate.
March 4th 2010, 01:22 AM
Archie Meade

For

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

Using the product rule of differentiation,

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

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

as

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

and

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

This is

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

or

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

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

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

When you asked, why is

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

not equal to

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

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

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