1. ## 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

2. Originally Posted by db5vry
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.

3. Originally Posted by Drexel28
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

4. Originally Posted by db5vry
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.

5. Originally Posted by Drexel28
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.

6. Originally Posted by db5vry
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.

7. Originally Posted by Drexel28
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

8. Originally Posted by db5vry
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.

9. Originally Posted by Drexel28
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

10. Originally Posted by db5vry
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.

11. Originally Posted by Drexel28
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

12. Originally Posted by db5vry
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) then we may conclude that it is a minimum, etc.

13. Originally Posted by Drexel28
$f(c)$ is by assumption either a local maximum or local minimum. So if $f(c) 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

14. Originally Posted by db5vry
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 $f'(x)$?

15. Originally Posted by Drexel28
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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