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Defferentiate this to simplest formy=a+bcosx/b+acosx

  1. Jun 20, 2004 #1
    can anybody defferentiate this to simplest form
  2. jcsd
  3. Jun 20, 2004 #2


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    Excuse me? bcosx/b= cos x doesn't it?
  4. Jun 20, 2004 #3
    I'm pretty sure he means

    [tex]y = \frac {a + b \cos x}{b + a \cos x}[/tex]

    The rule you want to use to differentiate this is

    [tex]{d\over dx} \left[{f(x)\over g(x)}\right]= {g(x)f'(x)-f(x)g'(x)\over [g(x)]^2}[/tex]

    This is called the "quotient rule". In your problem, f(x) = a + b cos x, and g(x) = b + a cos x. If a and b are constants, then f'(x) = -b sin x, and g'(x) = -a sin x. You can substitute those into the quotient rule and then simplify the result, and your problem is solved.

    Did that answer your question?
  5. Jun 21, 2004 #4
    i tried it but i am not gettting simplest form
  6. Jun 21, 2004 #5
    Do u mean Differential Eqn
  7. Jun 21, 2004 #6


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    Except for a pair of [tex] ~ab~sinx~cosx[/tex] terms that cancel off in the numerator, there is no other simplification to do.
  8. Jul 2, 2004 #7
    NOTE: Solution is wrong

    Okay here's a second method:

    [tex]y = \frac {a + b \cos x}{b + a \cos x}[/tex]

    Take log to the base e (ln) of both sides.

    ln(y) = ln(a + b \cos x) + ln(b+a \cos x)

    Differentiate both sides wrt x...

    \frac{1}{y}\frac{dy}{dx} = \frac{-b\sin x}{a+b\cos x} + \frac{-a}{b+a \cos x}

    Cross-multiply (or multiply both sides by y) and simplify...

    This method looks more tedious but is less prone to errors...the product rule can cause problems if the functions are complicated. But then there is no golden rule to say that one method is superior over the other..

    Hope that helps...


    Last edited: Jul 3, 2004
  9. Jul 2, 2004 #8


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    Maverick: You should have a minus sign between the two logarithms, not a plus sign.
  10. Jul 3, 2004 #9
    Oh yeah arildno...please pardon my mistake :-)

    There's also a mistake in the second term...which I have corrected here.

    Here's the correct portion:

    [tex]ln(y) = ln(a + b \cos x) + ln(b+a \cos x)[/tex]

    [tex]\frac{1}{y}\frac{dy}{dx} = \frac{-b\sin x}{a+b\cos x} + \frac{a\sin x}{b+a \cos x}[/tex]

    [NOTE: My previous solution is WRONG. Sorry for the inconvenience.]
  11. Jul 5, 2004 #10
    isn't SinX/CosX=TanX?
  12. Jul 5, 2004 #11
    by golly gosh, jimminy cricket there is right.
  13. Jul 6, 2004 #12
    come to think of it...y did u actually bring that into question? :confused:
  14. Jul 7, 2004 #13
    Yeah where does that come from in this question?
  15. Jul 8, 2004 #14
    :'( i was bored ok, and i wanted to know if it was right... sorry...
  16. Jul 8, 2004 #15
    well u were....but... yeah whatever it doesn't matter.

    hey...doesnt 2+2=4?
  17. Jul 10, 2004 #16
    This answer should be right.
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