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C3 Differentiation problem (quotient rule)

Could anyone help me with this problem where I've got stuck?

A curve has the equation y=lnxx2 y = \frac{lnx}{x^2} for x>0

I've got the first derivative which is 1−2lnxx3\frac{1-2 ln x}{x^3}

But the next part of the question is the following:

Find the coordinates of the maximum point of the curve and calculate the value of d2ydx2\frac{d^2y}{dx^2} there.

Having found the x-coordinate x=e12x = e^{\frac{1}{2}}, I'm having difficulty getting the y-coordinate... THis is what I've done:

y=lne12(e12)2y =\frac{ln e^\frac{1}{2}}{(e^\frac{1}{2})^2} = 12e14\frac{\frac{1}{2}} {e^\frac{1}{4}}

But I'm not sure how to proceed from here to get the value for y.

Any help would be much appreciated!
Reply 1
Avatar for joostan
joostan
13
Original post by jonnburton
Could anyone help me with this problem where I've got stuck?

A curve has the equation y=lnxx2 y = \frac{lnx}{x^2} for x>0

I've got the first derivative which is 1−2lnxx3\frac{1-2 ln x}{x^3}

But the next part of the question is the following:

Find the coordinates of the maximum point of the curve and calculate the value of d2ydx2\frac{d^2y}{dx^2} there.

Having found the x-coordinate x=e12x = e^{\frac{1}{2}}, I'm having difficulty getting the y-coordinate... THis is what I've done:

y=lne12(e12)2y =\frac{ln e^\frac{1}{2}}{(e^\frac{1}{2})^2} = 12e14\frac{\frac{1}{2}} {e^\frac{1}{4}}

But I'm not sure how to proceed from here to get the value for y.

Any help would be much appreciated!



A small error which may be the root of your troubles. . .:smile:
(e12)2≠e14(e^{\frac{1}{2}})^2 \not= e^{\frac{1}{4}}
(edited 10 years ago)
Reply 2
Avatar for jonnburton
jonnburton
OP
Thanks Joostan, it seems such an obvious mistake now that I think about what x12x^\frac{1}{2} actually means..!

So the value for y would be 12e \frac{\frac{1}{2}}{e} =12e\frac{1}{2e}
Reply 3
Avatar for joostan
joostan
13
Original post by jonnburton
Thanks Joostan, it seems such an obvious mistake now that I think about what x12x^\frac{1}{2} actually means..!

So the value for y would be 12e \frac{\frac{1}{2}}{e} =12e\frac{1}{2e}

Yep - A quote will illicit a faster response :tongue:
Reply 4
Avatar for jonnburton
jonnburton
OP
Original post by joostan
Yep - A quote will illicit a faster response :tongue:



OK thanks Joostan...



I have been trying to find the dydx2\frac{d^y}{dx^2} at the maximum point and keep running into problems.



I don't know if anyone can see where I'm going wrong:

1−2lnxx3\frac{1-2lnx}{x^3}

dydx2\frac{d^y}{dx^2}:

x3−2x2−x3∗2x−2x2∗2lnxx6\frac{x^3-2x^2 - x^3 *\frac{2}{x} - 2x^2 * 2ln x}{x^6}

= x3−2x2−x2−2x2∗2lnxx6 \frac{x^3 -2x^2 - x^2 -2x^2*2lnx}{x^6}

= x−2−1−2∗2lnxx4\frac{x -2-1- 2*2lnx}{x^4}

=x−3−4lnxx4\frac{x - 3 - 4lnx}{x^4}


putting in the value for x at the maximum point:

e12−3−4lne12e2 \frac{e^\frac{1}{2} - 3 - 4ln e^\frac{1}{2}}{e^2}

= e12−5e2\frac{e^\frac{1}{2} -5}{e^2}

But this doesn't get me closer to the book's answer, which is −2e2\frac {-2}{e^2}
Reply 5
Avatar for joostan
joostan
13
Original post by jonnburton
OK thanks Joostan...



I have been trying to find the dydx2\frac{d^y}{dx^2} at the maximum point and keep running into problems.



I don't know if anyone can see where I'm going wrong:

1−2lnxx3\frac{1-2lnx}{x^3}

dydx2\frac{d^y}{dx^2}:

x3−2x2−x3∗2x−2x2∗2lnxx6\frac{x^3-2x^2 - x^3 *\frac{2}{x} - 2x^2 * 2ln x}{x^6}

= x3−2x2−x2−2x2∗2lnxx6 \frac{x^3 -2x^2 - x^2 -2x^2*2lnx}{x^6}

= x−2−1−2∗2lnxx4\frac{x -2-1- 2*2lnx}{x^4}

=x−3−4lnxx4\frac{x - 3 - 4lnx}{x^4}


putting in the value for x at the maximum point:

e12−3−4lne12e2 \frac{e^\frac{1}{2} - 3 - 4ln e^\frac{1}{2}}{e^2}

= e12−5e2\frac{e^\frac{1}{2} -5}{e^2}

But this doesn't get me closer to the book's answer, which is −2e2\frac {-2}{e^2}


At a brief glance over your working it looks like you've differentiated incorrectly, that seems to be causing a problem later on. :smile:
It looks like you left the one in when differentiating the top, but as it's a constant it should disappear.
(edited 10 years ago)
Reply 6
Avatar for jonnburton
jonnburton
OP
I'm going to have to try it again! Cheers,
Original post by jonnburton
OK thanks Joostan...



I have been trying to find the dydx2\frac{d^y}{dx^2} at the maximum point and keep running into problems.



I don't know if anyone can see where I'm going wrong:

1−2lnxx3\frac{1-2lnx}{x^3}

dydx2\frac{d^y}{dx^2}:

x3−2x2−x3∗2x−2x2∗2lnxx6\frac{x^3-2x^2 - x^3 *\frac{2}{x} - 2x^2 * 2ln x}{x^6}

= x3−2x2−x2−2x2∗2lnxx6 \frac{x^3 -2x^2 - x^2 -2x^2*2lnx}{x^6}

= x−2−1−2∗2lnxx4\frac{x -2-1- 2*2lnx}{x^4}

=x−3−4lnxx4\frac{x - 3 - 4lnx}{x^4}


putting in the value for x at the maximum point:

e12−3−4lne12e2 \frac{e^\frac{1}{2} - 3 - 4ln e^\frac{1}{2}}{e^2}

= e12−5e2\frac{e^\frac{1}{2} -5}{e^2}

But this doesn't get me closer to the book's answer, which is −2e2\frac {-2}{e^2}


An easier way to find the second derivative is as follows
dydx=1−2ln⁡xx3  ⟹  x3dydx=1−2ln⁡x \frac{dy}{dx}=\frac{1-2\ln{x}}{x^3} \implies x^3\frac{dy}{dx}=1-2\ln{x}
Then differentiate, substitute for dydx\frac{dy}{dx} and rearrange to get the second derivative

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