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Math Help - slope of tangent line to a curve

  1. #1
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    slope of tangent line to a curve

    ok so here's my question
    Find the slope of the tangent line to the curve x^y = y^x at the point (2,4)

    im not sure exactly how to go about this, what i've tried so far is taking the natural log of both sides so move the exponents
    lnx^y=lny^x
    ylnx = xlny
    and that's where i get stuck. Am I supposed to differentiate both sides then? Because when I did that so that then I could isolate y' i got the following


    y' = y(y-xlny)/x(x-ylnx)
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  2. #2
    Super Member TheChaz's Avatar
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    Good! Now just put 2 in for x, and 4 in for y.
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  3. #3
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    Ok so after doing that what do I do? I'm not allowed to use a calculator on this problem so I can't get an exact number, would I leave it simply as
    y' = 4(4-2ln4)/2(2-4ln2)
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  4. #4
    Super Member TheChaz's Avatar
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    First, let me disclaim: I haven't actually calculated anything - I'm just helping you with the concept under the assumption that you know the product and chain rules!

    The fraction you have can be reduced by 2 on top and bottom. Other than that... I can't come up with any reasonable ways to rewrite. Then again, I'm on my iPhone...
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  5. #5
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    Haha that's alright I am SUPPOSED to know how to do this already, but usually we get easier problems like y = x^2-3x or something like that.... i know that for that you would plug in the x in order to get f(a) then take the derivative and set it equal to zero which gives the slope m, then you just use point slope formula y - f(a) = m (x - a)

    i'm just having trouble applying that concept to this problem primarily because I don't have it written in terms of y = something you know?
    and i don't know if i'm supposed to actually set that derivative that i got, assuming that it's correct, and setting it equal to zero. I know that for this problem a zero would occur where the numerator is zero, so should I set the numerator equal to zero an solve? Mmmmmm I'm just confusing myself :/
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  6. #6
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    Why would you set the derivative equal to 0? You're not evaluating stationary points.

    You want to know what the derivative is at the point \displaystyle (x,y) = (2, 4).

    You have an \displaystyle x,y relationship to give you the derivative at any point, so substitute \displaystyle x = 2 and \displaystyle yt = 4.


    Once you have the derivative at that point, you can use it and the \displaystyle (x,y) co-ordinate to find the equation of the tangent line using \displaystyle y = mx + c...
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  7. #7
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    Are you asking me why I would set it equal to zero in general, or why would I set it equal to zero in this particular case? In this case I didn't know if I should as I am not familiar with this problem, and my unfamiliarity with it would be the reason I would attempt it. Either way I was thinking too much about the problem. Question asks for slope, which I have found, it says nothing about the equation of a line tangent at that point.
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  8. #8
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    You found y' = \dfrac{4(4-2\ln4)}{2(2-4\ln2)}, which is the slope of the line tangent to the curve at the point (2,4). Here is how I would simplify the answer:

    y' = \dfrac{4(4-2\ln(2^2))}{2(2-4\ln2)}

    \implies y' = \dfrac{4(4-4\ln2)}{2(2-4\ln2)}

    \implies y' = \dfrac{16(1-\ln2)}{4(1-2\ln2)}

    \implies y' = \dfrac{4(1-\ln2)}{(1-2\ln2)}

    If you aren't allowed to use a calculator, then I assume the above answer is acceptable. Any decimal approximation would be just that - an approximation.
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