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Math Help - can someone explain to me..

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by tbenne3 View Post
    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
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  3. #3
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    Quote Originally Posted by Archie Meade View Post
    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)
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  4. #4
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    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!
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  5. #5
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    Quote Originally Posted by experiment00005 View Post
    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)
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  6. #6
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    Quote Originally Posted by tbenne3 View Post
    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
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  7. #7
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    Quote Originally Posted by tbenne3 View Post
    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?
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  8. #8
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    Quote Originally Posted by Kasper View Post
    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?
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  9. #9
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    Quote Originally Posted by tbenne3 View Post
    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.
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  10. #10
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    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)
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