Use the chain rule or product rule?

Discussion in 'Undergraduate Math' started by trixx, Dec 5, 2008.

  1. trixx

    trixx Guest

    I'm trying to find the derivative of this trig function, and I used the product rule instead of the chain rule, so obviously my book had a different answer.

    Here's part of the problem:

    differentiate tanx * sqrt(sinx)

    the product rule gave me:

    sqrt(sinx)sec²x + cosx/2sqrt(sinx) tanx

    the book used the chain rule and got:

    sec²x sqrt(sinx) * cosx/2sqrt(sinx)

    I'm wondering if they're the same...they should be, right? I obviously used the chain rule to get the sqrt(sinx).

    Thanks in advance!
     
    trixx, Dec 5, 2008
    #1
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  2. This requires you to use the product rule first, and then to use the
    Chain rule to find the derivative of sqrt(sin x).
    This is imprecise. You mean

    sqrt(sin x)sec^2(x) + cos x*tan x/2sqrt(sin x)

    (otherwise, it looks like "tan x" is multiplying 2*sqrt(sin x) in the
    denominator).
    What do you mean, "they used the Chain Rule"?

    Can you provide the steps they used? This is not even simplified, so I
    suspect you copied this incorrectly. You would be able to cancel
    sqrt(sin x) which appears in both numerator and denominator.
    If you copied everything correctly, perhaps. Please give the steps,
    and be sure to use enough parenthesis so there is no ambiguity in what
    you write.

    --
    ======================================================================
    "It's not denial. I'm just very selective about
    what I accept as reality."
    --- Calvin ("Calvin and Hobbes" by Bill Watterson)
    ======================================================================

    Arturo Magidin
    magidin-at-member-ams-org
     
    Arturo Magidin, Dec 5, 2008
    #2
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  3. trixx

    trixx Guest

    I rewrote my question on paper and scanned it, I think it'll be much clearer. I also think I figured out the answer to my question, but I want to be sure.

    http://img254.imageshack.us/img254/3752/mathfullxz3.jpg

    I should never have used the product rule on that problem, because tan(sqrt(sin)) is one function inside of another. I must've confused it for tan(x) * sqrt(sinx) in which case the product rule would've been the right choice.
     
    trixx, Dec 5, 2008
    #3
  4. Please quote the message you are replying to to provide context. Most
    people do not use the Math Forum interface to read this newsgroup.
    "tan(sqrt(sin))" is not a function. There is no variable. You meant
    "tan(sqrt(sin(x)))".

    Being so cavalier with function names and variables is precisely what
    leads to mistakes, like thinking that you have a product when in fact
    you have a composition.
    Actually, it is neither! Sigh.

    Your scanned page has the function

    sin (tan ( sqrt (sin x))).

    And yes, this is something completely different from a product, which
    is why you had an answer different from the book.

    The derivative of this function (which is neither tan(sqrt(sin x)) nor
    tan(x)*sqrt(sin(x))), is:

    f(x) = sin( tan (sqrt (sin x))),

    f'(x) = cos(tan(sqrt(sin(x))))*sec^2(sqrt(sin(x)))*cos(x)/2sqrt(sin(x)).

    What you have marked as "inside derivative found using the Product
    Rule and the Chain Rule" is wrong, based on you misunderstanding the
    notation: tan(sqrt(sin(x))) is not tan(x)*sqrt(sin(x)). That mistake
    is born from omitting the variable when writing "tan", "sin", "cos",
    etc. Always include the variable. So, the answer to the question at
    the bottom of the page is "no", they are not equal, because you are
    taking two different functions.


    --
    ======================================================================
    "It's not denial. I'm just very selective about
    what I accept as reality."
    --- Calvin ("Calvin and Hobbes" by Bill Watterson)
    ======================================================================

    Arturo Magidin
    magidin-at-member-ams-org
     
    Arturo Magidin, Dec 5, 2008
    #4
  5. trixx

    Passerby Guest

    FWIW here's a web site you can use to perform derivatives step-by-step
    (among other things) to check your work.
    <http://www.calc101.com/>
     
    Passerby, Dec 5, 2008
    #5
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