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    Question:

    given

    y=\dfrac{4\ln(x-3)}{4\ln(x+3)}

    show that

    \dfrac{dy}{dx}=\dfrac{24}{x(4\ln  (x+3))}

    __________________________-



    I can't seem to get there. Cancelling the 4's and using the quotient rule gives me

    \dfrac{dy}{dx}=\dfrac{1}{(x-3)\ln(x+3)}-\dfrac{\ln(x-3)}{(x+3)(\ln(x+3))^2}

    which doesn't look much like what they've got
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    I get the same answer as you, what do they have?
    • Thread Starter
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    (Original post by jameswhughes)
    I get the same answer as you, what do they have?
    they have what I posted. "show that"
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    (Original post by Plato's Trousers)
    they have what I posted. "show that"
    oops, didn't see that. i can't get it either :/
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    They've made a mistake I think, try plugging in values
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    try using the product rule instead.
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    I haven't used the qoutient rule in a few weeks, i guess il have to go back to it and recap.
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    Community Assistant
    Ewan has already sorted this one.
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    Use implicit differentiation, works for me
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    (Original post by Scottishsiv)
    Use implicit differentiation, works for me
    Could you explain please?
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    (Original post by Scottishsiv)
    Use implicit differentiation, works for me
    There is nothing implicit to differentiate here - everything's explicit unless you're trying to turn something that is explicit into an implicit form, but that almost never helps.
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    (Original post by Farhan.Hanif93)
    There is nothing implicit to differentiate here, everything's explicit.
    I think he meant you could multiply both sides by the denominator.

    However substitution shows the answer Plato has got (which is correct) is not equivalent to the supposedly simplified one anyway.
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    Well first turn it into:

    yln(x+3) = ln(x-3)

    then differentiate both sides using implicit differentiation

    the you get dy/dx ln(x+3) = 1/(x-3) - y/(x+3)

    then sub in y = ln(x-3) / ln (x+3)

    I think that works, haven't worked further than this though. apologies for the layout.
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    (Original post by Mr M)
    I think he meant you could multiply both sides by the denominator.

    However substitution shows the answer Plato has got (which is correct) is not equivalent to the supposedly simplified one anyway.
    Yep, edited my post just as you were typing.
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    (Original post by Scottishsiv)
    Well first turn it into:

    yln(x+3) = ln(x-3)

    then differentiate both sides using implicit differentiation

    the you get dy/dx ln(x+3) = 1/(x-3) - y/(x+3)

    then sub in y = ln(x-3) / ln (x+3)

    I think that works, haven't worked further than this though. apologies for the layout.
    From which it follows that you get what Plato got, which is the right answer but it doesn't match the given answer.
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    thanks everyone! Reassuring
 
 
 
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