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OK who wants to do a calculus problem?

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  1. #1

    Default OK who wants to do a calculus problem?

    Find dy/dx (or just call it y') by implicit differentiation.

    tan(x/y) = x + y

    I see the answer but I need to understand it. Can someone explain it well plz? I did the following so far:

    1) Derivative of tan(x/y) = [sec(x/y)]^2 * [(y-xy')/y^2]
    is equal to
    2) Derivative of x + y = 1 + y'

    Is this right so far? If so then I'm just having trouble simplifying the way the book does I guess.
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  2. #2
    gabe's Avatar
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    what exactly is given (not the answer), and what exactly do you need to show
  3. #3
    gabe's Avatar
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    "Is this right so far?"
    yes
  4. #4
    Quote Originally Posted by gabe
    what exactly is given (not the answer), and what exactly do you need to show
    tan(x/y) = x + y is given

    answer in back of book is...

    y' = [ysec^2(x/y)-y^2] / [y^2 + xsec^2(x/y)]
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  5. #5
    Quote Originally Posted by martindcx1e
    Quote Originally Posted by gabe
    what exactly is given (not the answer), and what exactly do you need to show
    tan(x/y) = x + y is given

    answer in back of book is...

    y' = [ysec^2(x/y)-y^2] / [y^2 + xsec^2(x/y)]
    Are you sure you typed that in correctly because I'm getting the same answer except with the bold term not there at all.
  6. #6
    i typed it correctly mcat
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  7. #7
    gabe's Avatar
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    so all you need to do is solve for y' in the equatino you wrote in (1)

    [sec(x/y)]^2 * [(y-xy')/y^2] = 1 + y'

    so distribute the left side, then move the term with the y' in it to the right side

    ill post it out if that doesnt help
  8. #8
    never mind their answer is correct

    try setting that funky secant expression to equal K and then simplify it that'll make things easier
  9. #9
    ah thanks mcat that worked out nicely
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  10. #10
    MATH FUCK YEAH
  11. #11
    lolzzz_321's Avatar
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    what you know bout math?

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