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Differentiate Trigo

  1. Nov 12, 2009 #1
    18tan^2(x)sec^2(x)

    Is this correct?

    Product Rule I think...

    18tan^2(x)sec^2(x) = (18tan^2(x))(2sec^3(x)) + (sec^2(x))(36tan^3(x)) = (36tan^3(1+tan^2(x))) + (36tan^3(1+tan^2(x))) = 2(36tan^3(1+tan^2(x)))

    1 more thing is tan x = sec^2(x)? Any 1 has a list of Trigo formulas? Becoz' I am stump here as my course assumes we learned all this stuff from High School, but I didn't attend High School...
     
  2. jcsd
  3. Nov 12, 2009 #2
    tan(x) is not sec^2(x).

    Can you please state your understanding of what the product rule says, what the chain rule says, what the derivative of tan(x) is, and what the derivative of sec(x) is? You're making an error in your calculations, and I want to know where the misunderstanding is or if you just made an arithmetic mistake.
     
  4. Nov 12, 2009 #3
    Erm... the product rule is when you have 2 parts of an equation like (x)(y), so to different them... u take (x)(y)' + (x)'(y) when ' denotes differentiate it?

    Derivative of tan x... I have no idea how to calculate it, but formula say sec^2(x)?
     
  5. Nov 12, 2009 #4
    Yes, the derivative of tan x is sec^2 x, which can be done using tan x = sin x / cos x and doing a product/quotient rule.

    Your understanding of the product rule looks good.

    In your first step, then you evaluate the derivative of sec^2(x) to be 2sec^3(x)? How did you arrive at that?
     
  6. Nov 12, 2009 #5
    Meh... I got it mixed up with integration... its supposed to be 2sec(x) right?
     
  7. Nov 12, 2009 #6
    ...and then you proceed with the chain rule.

    the derivative of f(g(x)) is f'(g(x))*g'(x)
     
  8. Nov 12, 2009 #7
    Okie. Just one more question, I am sure I am right this time

    I've been told to use chain rule to solve this 2+tan^3

    So I take 2+tan^3 as my U, where after differentiate I get a 1 because anything to the power of 0 is 1.

    The taking 1 x (2+tan^3(x))' = 3tan^2(x)... but it seems my answer is wrong? I am suppose to get 3tan^2(x)(tanx)' or (3tan^2)(sec^2(x))?

    And how do u integral U^-1? U can't power rule...
     
    Last edited: Nov 12, 2009
  9. Nov 12, 2009 #8

    Mark44

    Staff: Mentor

    Your use of terminology is pretty sloppy and unclear; for instance "solve 2 + tan^3" is totally meaningless. I'm assuming that you want to differentiate 2 + tan3(x).

    If so, then d/dx(2 + tan3(x)) = d/dx(2) + d/dx(tan3(x)), using the sum rule.

    Now use the chain rule to get d/dx(tan3(x)).

    Why do you want to integrate? The power rule for integrals says that
    [tex]\int u^n du~=~ \frac{u^{n + 1}}{n + 1}~for~n~\neq~-1[/tex]

    There's a different formula to use if n = -1.
     
  10. Nov 12, 2009 #9
    Question for d/dx(tan3(x)) <- Y chain rule? There is only 1 term and a power why is it we can't use power rule?

    Secondly, whats the different formula for n = -1?

    Sorry if I been asking alot of stupid questions but me and my pals have been thrown into taking this module without any prior foundation knowledge, we don't know Trigo, angles and everything. And the teachers at collage only touch on stuff briefly. I've been rewatching lecture videos over and over again and they still didn't explain anything. So I been searching the net for stuff. Sorry again if I asked too much.
     
  11. Nov 12, 2009 #10
    You have to use the chain rule for tan^3(x) because you have two functions there:
    f(u) = u^3
    g(x) = tan(x)

    tan^3(x) is (tan(x))^3, which is f(g(x)).

    When you take the derivative of tan^3(x), you have to take the derivative of u^3 with respect to u, which gives 3u^2, then multiply by the derivative of tan(x), which is sec^2(x). Then substitute back in for u.

    I would recommend approaching your teacher about this stuff. If you've never done trigonometry and you're having trouble with function composition, perhaps you're taking the wrong class.
     
  12. Nov 12, 2009 #11
    So we dun view it as tan^3(x) like x^3? So u mean tan^3(x) is different from tanX^3?

    So wads the formula for n=-1 for Integration, I can't seen find it. I have this du/u^2 but I am kinda stuck there.

    Nah, its the education system over here. There's a split during our so called High School, where students get the choose to continue High School or go for vocational stuff. The high school usually prepares them for collage, however high scorers like 15% of the ppl in the vocational school are given a chance to enter too. However uni is pretty much catered to the high school ppl, so when it comes to maths most of us suffer.
     
  13. Nov 12, 2009 #12

    Mark44

    Staff: Mentor

    tan3(x) is different from tan(x3). The first is notation that really means (tan(x))3, or (tan(x)*tan(x)*tan(x)). The second means take the cube of x, then take the tangent of the result.
    The two formulas relating to the power rule for integration are
    [tex]\int u^n du~=~ \frac{u^{n + 1}}{n + 1} + C~for~n~\neq~-1[/tex]

    and
    [tex]\int u^{-1} du~=~ln|u| + C [/tex]

    To integrate 1/u2, rewrite this as u-2 and use the first formula above.
     
  14. Nov 12, 2009 #13
    Hey I saw that! But the correct answer I have is -(1/U) + C how did that happen? Where's the In and stuff?

    The initial part was U/(U^2) then after integration the answer is (-1/U) + C, no In?
     
  15. Nov 12, 2009 #14

    Mark44

    Staff: Mentor

    That's Ln, not In. Ln stands for natural logarithm (in Latin, with the order reversed).
    [tex]\int \frac{du}{u^2}~=~\int u^{-2} du~=~\frac{u^{-1}}{-1} + C~=~\frac{-1}{u} + C[/tex]

    It is NOT true that the integral of du over whatever is Ln|whatever|. It's important to understand the "fine print" in the power rule for integrals; when you can apply it and when you can't.
     
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