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Y=tanx y'=sec^2x does y' also equal to (secx)^2 ?

  1. Oct 6, 2004 #1
    y=tanx
    y'=sec^2x

    does y' also equal to (secx)^2 ?
     
    Last edited: Oct 6, 2004
  2. jcsd
  3. Oct 6, 2004 #2

    arildno

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    [tex]sec^{2}x\equiv(sec{x})^{2}[/tex]
    The first is simply a fancy way of writing the square of secx
     
  4. Oct 6, 2004 #3

    Diane_

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    Yes. sec^2x is just the standard way of writing (sec x)^2. It helps avoid confusion with sec (x^2).
     
  5. Oct 6, 2004 #4
    my text book also has something about
    d/dx a^u = a^u ln a du/dx

    what's that about?
     
  6. Oct 6, 2004 #5

    arildno

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    That's a chain rule use, with "a" a constant, "u" a function of "x".
     
  7. Oct 6, 2004 #6

    Pyrrhus

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    When you have something like

    [tex] y = a^u [/tex]

    you can solve it as

    [tex] ln y = ln a^u [/tex]

    [tex] ln y = u ln a [/tex]

    Derivating the above we will have

    [tex] \frac{y'}{y} = u' ln a [/tex]

    [tex] y' = y u' ln a [/tex]

    [tex] y' = a^u ln a u'[/tex]
     
  8. Oct 6, 2004 #7
    I thought chain rule was something like:
    d/dx f(q(x))=f'(g(x))*q'(x)
     
  9. Oct 6, 2004 #8

    Diane_

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    It can be derived using the fundamental rules.

    Let a^u = y

    ln (a^u) = ln(y)

    u ln(a) = ln(y)

    d/dx(u ln(a)) = d/dx (ln(y))

    For the first part, remember the product rule:

    d/dx(u ln(a)) = u d/dx(ln(a)) + ln(a) d/dx(u)

    Since a is a constant, d/dx(ln(a)) = 0, so

    d/dx(u ln(a)) = ln(a) du/dx

    For the other side, use the chain rule:

    d/dx (ln(y)) = 1/y dy/dx

    So:

    ln(a) du/dx = 1/y dy/dx

    dy/dx = y ln(a) du/dx

    Remember that y = a^u, so

    d/dx(a^u) = a^u ln(a) du/dx

    QED = quod erat demonstrandum = quite easily done. :)
     
  10. Oct 6, 2004 #9

    arildno

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    That's right:
    [tex]\frac{d}{du}a^{u}=a^{u}ln(a)[/tex]
    Hence,
    [tex]\frac{d}{dx}a^{u(x)}=(\frac{d}{du}a^{u})\frac{du}{dx}=a^{u(x)}ln(a)\frac{du}{dx}[/tex]
     
  11. Oct 6, 2004 #10

    Pyrrhus

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    Bah, you both did it more fancy :cry:
     
  12. Oct 6, 2004 #11
    so

    d/dx (a^2+2)^2=2(a^2+2)*2a
    but it also equals
    d/dx (a^2+2)^2=(a^2+2)^2*ln(a^2+2)*2a
    ???
     
  13. Oct 6, 2004 #12

    arildno

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    That's what you've gotta live with when you deliver too speedy answers
    ..:biggrin:
     
  14. Oct 6, 2004 #13

    arildno

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    WHAT?????????
    2 is not a varying function, is it?
    Besides, shouldn't it be d/da???
     
  15. Oct 6, 2004 #14
    :rofl:

    I guess I was confused. Can you give an example of where the derivative of a^u can be used? As in show me an easy example problem?
     
  16. Oct 6, 2004 #15

    arildno

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    I'm not following..used???
     
  17. Oct 6, 2004 #16
    As in do a sample problem, like how I made up the equation (a^2+2)^2 and found the derivative with the chain rule: d/dx (a^2+2)^2=2(a^2+2)*2a
     
  18. Oct 6, 2004 #17

    Pyrrhus

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    He means a function, like [tex] f(x) = 3^x [/tex]
     
  19. Oct 6, 2004 #18
    ohhhh, I understand totally what it is used for now.... the "a" has to be a number without a variable!
     
  20. Oct 7, 2004 #19

    HallsofIvy

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    [tex] f(x)= 2^{x}[/tex]

    [tex] \frac{df}{dx}= (ln 2)2^x [/tex]

    More generally,

    [tex] f(x)= 3^{x^2- 3x+ 2}[/tex]

    [tex]\frac{df}{dx}= (2x- 3)(ln 3)3^{x^2- 3x+ 2}[/tex]
     
  21. Oct 7, 2004 #20

    Diane_

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    Here's to pure mathematics - may it never be of any use to anyone. :)
     
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