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Math Help - Logarithmic differentiation

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    Logarithmic differentiation

    Given that f(x) = x^{-\sqrt x} for  x > 0,

    (a) obtain an expression for f′ (x), [4]
    (b) find the x-coordinate of the stationary point on the graph of f and determine whether this point is a maximum or a minimum. [4]

    Now for the first part, I obtained a derivative of

    (x^{-\sqrt x}) \frac{-1}{\sqrt x} - \frac{lnx}{2\sqrt x} through logarithmic differentiation. I think this is correct, but the problem I have is knowing what to do in the second part of the question. How would you go about finding the x-coordinate and how can you tell what kind of point it is?

    Thanks again for your time and patience
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by db5vry View Post
    Given that f(x) = x^{-\sqrt x} for  x > 0,

    (a) obtain an expression for f′ (x), [4]
    (b) find the x-coordinate of the stationary point on the graph of f and determine whether this point is a maximum or a minimum. [4]

    Now for the first part, I obtained a derivative of

    (x^{-\sqrt x}) \frac{-1}{\sqrt x} - \frac{lnx}{2\sqrt x} through logarithmic differentiation. I think this is correct, but the problem I have is knowing what to do in the second part of the question. How would you go about finding the x-coordinate and how can you tell what kind of point it is?

    Thanks again for your time and patience
    Solve for zero. Use the second derivative test.
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    Quote Originally Posted by Drexel28 View Post
    Solve for zero. Use the second derivative test.
    So would I take \frac{d^2y}{dx^2} of the derivative I obtained?

    It seems a very complicated function and I would find that enormously difficult to even attempt! Would you describe it as a lengthier process than the first, because both parts of the question are worth 4 marks each and the second part sounds like a higher volume of work
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by db5vry View Post
    So would I take \frac{d^2y}{dx^2} of the derivative I obtained?

    It seems a very complicated function and I would find that enormously difficult to even attempt! Would you describe it as a lengthier process than the first, because both parts of the question are worth 4 marks each and the second part sounds like a higher volume of work
    Plug in local values. Be careful of this method though. Or, alternatively "cheat" and just graph it. Once again though, I would advise using the second derivative test or the slightly more innocuous first derivative test.
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    Quote Originally Posted by Drexel28 View Post
    Plug in local values. Be careful of this method though. Or, alternatively "cheat" and just graph it. Once again though, I would advise using the second derivative test or the slightly more innocuous first derivative test.
    I like your idea of plugging in a value I couldn't cheat as this would be for an exam which doesn't allow graphic calculators - on my course syllabus it says nothing about taking second derivatives (It is an A-Level course for further maths) so I assume that plugging in values must be what they are asking for

    I plugged in x=1 as I recognised that ln 1 = 0 so thought that would automatically make the rest of it zero - but then I checked it and it did not mean this was the case, as it is only being subtracted from something else.

    I'm struggling here but I hope I can learn how to answer it using your recommended method of second derivatives - I can only do this with basic/parametric equations but f'(x) in this question looks beyond my ability to differentiate a second time.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by db5vry View Post
    I like your idea of plugging in a value I couldn't cheat as this would be for an exam which doesn't allow graphic calculators - on my course syllabus it says nothing about taking second derivatives (It is an A-Level course for further maths) so I assume that plugging in values must be what they are asking for

    I plugged in x=1 as I recognised that ln 1 = 0 so thought that would automatically make the rest of it zero - but then I checked it and it did not mean this was the case, as it is only being subtracted from something else.

    I'm struggling here but I hope I can learn how to answer it using your recommended method of second derivatives - I can only do this with basic/parametric equations but f'(x) in this question looks beyond my ability to differentiate a second time.
    Why is the second derivative so hard. Let f(x)=x^{-\sqrt{x}}. I made a slight clerical error. That being said, why don't you try the second derivative? You can validate it here wolframalpha.com. It will even tell you how to compute the derivative.
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    Quote Originally Posted by Drexel28 View Post
    Why is the second derivative so hard. Let f(x)=x^{-\sqrt{x}}. I made a slight clerical error. That being said, why don't you try the second derivative? You can validate it here wolframalpha.com. It will even tell you how to compute the derivative.
    I actually used that site to validate the first derviative that I took and I had a look to see what the second one would be, but that is far beyond my capability which is why I find it difficult I did try, and took logs of both sides and I couldn't find a way to get any further. Sorry but I'm not sure the question is looking for the answer through those methods - maybe the plugging in the values that was suggested? How would you go about doing that? I have never done this before and I would like to know how
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    Quote Originally Posted by db5vry View Post
    I actually used that site to validate the first derviative that I took and I had a look to see what the second one would be, but that is far beyond my capability which is why I find it difficult I did try, and took logs of both sides and I couldn't find a way to get any further. Sorry but I'm not sure the question is looking for the answer through those methods - maybe the plugging in the values that was suggested? How would you go about doing that? I have never done this before and I would like to know how
    This method is EXTREMELY sketchy. The concept is this. Having found a critical point c of f(x) we can assume (by the way the question is worded) that f(c) is either a local maximum or local minimum. To determine which it is take f(c) and compare it to f\left(c+\varepsilon\right) for some arbitrarily small \varepsilon. The sketchiness is that if you use \varepsilon=\tfrac{1}{2} was it small enough.
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    Quote Originally Posted by Drexel28 View Post
    This method is EXTREMELY sketchy. The concept is this. Having found a critical point c of f(x) we can assume (by the way the question is worded) that f(c) is either a local maximum or local minimum. To determine which it is take f(c) and compare it to f\left(c+\varepsilon\right) for some arbitrarily small \varepsilon. The sketchiness is that if you use \varepsilon=\tfrac{1}{2} was it small enough.
    I see what you mean. I've tried this value and also some smaller values and none of them come close to equating to zero. The question seems to be looking for a very particular answer and it sounds definitely very hit and miss.

    The question is on this past paper, just for you to get a bit of background:

    http://www.wjec.co.uk/uploads/papers/w08-977-01.pdf it's the ninth and last question on the paper. I just have this feeling that I'm missing something completely obvious but I don't know what it is
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by db5vry View Post
    I see what you mean. I've tried this value and also some smaller values and none of them come close to equating to zero. The question seems to be looking for a very particular answer and it sounds definitely very hit and miss.

    The question is on this past paper, just for you to get a bit of background:

    http://www.wjec.co.uk/uploads/papers/w08-977-01.pdf it's the ninth and last question on the paper. I just have this feeling that I'm missing something completely obvious but I don't know what it is
    You seem to have used the method I described to find a zero of f'(x). Instead, it is used to determine, when found, whether that zero admits a maximum or minimum value.
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    Quote Originally Posted by Drexel28 View Post
    You seem to have used the method I described to find a zero of f'(x). Instead, it is used to determine, when found, whether that zero admits a maximum or minimum value.
    I now interpret what you mean as inserting a small value as x and seeing whether it is positive or negative - I put in 0.2 and got a negative value, which would mean that it is a maximum point?

    Sorry if I have misunderstood again, it must be frustrating but I massively appreciate your help
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    Quote Originally Posted by db5vry View Post
    I now interpret what you mean as inserting a small value as x and seeing whether it is positive or negative - I put in 0.2 and got a negative value, which would mean that it is a maximum point?

    Sorry if I have misunderstood again, it must be frustrating but I massively appreciate your help
    f(c) is by assumption either a local maximum or local minimum. So if f(c)<f\left(c+\varepsilon\right) then we may conclude that it is a minimum, etc.
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    Quote Originally Posted by Drexel28 View Post
    f(c) is by assumption either a local maximum or local minimum. So if f(c)<f\left(c+\varepsilon\right) then we may conclude that it is a minimum, etc.
    Well if I called f(c) f(0.2), I obtained a value of -0.89 to 2dp.
    If I called f(E) f(0.02), I obtained 11.75.

    The smaller value has seen the sign change. I didn't think that would have happened but I'm sure it's incorrect
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by db5vry View Post
    Well if I called f(c) f(0.2), I obtained a value of -0.89 to 2dp.
    If I called f(E) f(0.02), I obtained 11.75.

    The smaller value has seen the sign change. I didn't think that would have happened but I'm sure it's incorrect
    Is .2 the zero of f'(x)?
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    Quote Originally Posted by Drexel28 View Post
    Is .2 the zero of f'(x)?
    It isn't, I am not sure how to do that at all. I feel increasingly out of my depth with this question
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