Logarithmic differentiation

**db5vry** · January 2nd 2010, 08:32 PM

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

**Drexel28** · January 2nd 2010, 08:34 PM

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.

**db5vry** · January 2nd 2010, 08:42 PM

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

**Drexel28** · January 2nd 2010, 08:45 PM

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.

**db5vry** · January 2nd 2010, 08:54 PM

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.

**Drexel28** · January 2nd 2010, 09:24 PM

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.

**db5vry** · January 2nd 2010, 09:39 PM

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

**Drexel28** · January 2nd 2010, 09:43 PM

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.

**db5vry** · January 2nd 2010, 09:55 PM

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

**Drexel28** · January 2nd 2010, 09:58 PM

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.

**db5vry** · January 2nd 2010, 10:05 PM

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

**Drexel28** · January 2nd 2010, 10:07 PM

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 $f(c)<f\left(c+\varepsilon\right)$ then we may conclude that it is a minimum, etc.

**db5vry** · January 2nd 2010, 10:15 PM

Originally Posted by Drexel28

$f(c)$ 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

**Drexel28** · January 2nd 2010, 10:22 PM

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 $f'(x)$ ?

**db5vry** · January 2nd 2010, 10:26 PM

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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