Logs Differentiation/integration help! (1 Viewer)

[0hNivlek]

Hey guys, it's me again! As you may have noticed, maths is like my worst subject lol. Btw, these are just the questions in the excerices i couldn't get ;x!

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Thanks for you help guys!

 

[lolokay]

1. just use chain rule as per usual

2. differentiate with product rule, equate to zero

3. simplify first using log law

4. divide each term by x² and integrate each individually

 

[LordPc]

Id like to help but I cant quite use this LaTeX tool

like how do i go to the next line?

 

[tommykins]

use \\

 

[0hNivlek]

Thanks for the quick replies ;D! I just re-did question 1 and 3 and i got it now w00t. Thanks!

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[lolokay]

your working is correct so far
but what are you trying to solve for? (what is the question asking)

 

[AlexJB]

Thanks for the quick replies ;D! I just re-did question 1 and 3 and i got it now w00t. Thanks!

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Question 2 has to be wrong. The derivative of xlogx = logx + 1. Can't have a negative value for x. A stationary point exists at 1/e but not -1/e

 

[lolokay]

^ read the question again

 

[AlexJB]

y = xlogx. Differentiating that gives y' = log x + 1. Equating that to 0 gives logx + 1 = 0, logx = -1. Using the rule:

y = logax
x = a ^ y

gives: x = e ^ -1 = 1 / e.
y'' = 1/x = pos. Therefore min point. So a min.point exists at 1/e.

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I read it again, I did it again, same result. They're saying, show the minimum point of y = xlogx = -1/e.

y' = logx + 1
= log(-1/e) + 1
= math error.

It isn't a real solution, therefore it can't be a min.

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Edit: for question four, just divide all by x^2.






That should allow you to do it pretty comfortably.

 

[lolokay]

.......

Spoiler

the minimum occurs at x = 1/e
y = xlnx
= 1/e ln(1/e) at the minimum
= 1/e *-1
= -1/e

 

[Timothy.Siu]

y = xlogx. Differentiating that gives y' = log x + 1. Equating that to 0 gives logx + 1 = 0, logx = -1. Using the rule:

y = logax
x = a ^ y

gives: x = e ^ -1 = 1 / e.
y'' = 1/x = pos. Therefore min point. So a min.point exists at 1/e.

---

I read it again, I did it again, same result. They're saying, show the minimum point of y = xlogx = -1/e.

y' = logx + 1
= log(-1/e) + 1
= math error.

It isn't a real solution, therefore it can't be a min.

y'=ln x +1
y'=0 when x=1/e

x=1/e y=-1/e

 

[tommykins]

Minimum VALUE, not point.

lolokay is correct.

 

[Timothy.Siu]

.......

Spoiler

the minimum occurs at x = 1/e
y = xlnx
= 1/e ln(1/e) at the minimum
= 1/e *-1
= -1/e

lol u beat me

 

[0hNivlek]

Thanks guys. I solved all those problems now ;D!

Now i got a new dilemna! In the equation below, i need help integrating the 4/x^2 part. I can find the volume and stuff, but i just don't understand the integration part. I mean all the examples/questions follow the f'(x)/f(x) = ln (f(x)) + c rule. So how do i integrate 4/x^2?



Thanks!@

 

[Trebla]

Do you know how to integrate ?

 

[0hNivlek]

= 4x^-2
=4x^-2+1 / -1
=4x^-1 / -1
=-4/x

Yup. Don't i need to use logs to do it?

 

[lolokay]

well, you just did it without logs, so...

 

[Trebla]

= 4x^-2
=4x^-2+1 / -1
=4x^-1 / -1
=-4/x

Yup. Don't i need to use logs to do it?

First of all, 4x^-2 is NOT equal to 4x^-2+1 / -1. What you've just done was integrate it already lol

 

[0hNivlek]

^ thanks for the help. I finally solved that one LOL.

Gosh. Now i have another dilemna.

In the Jones&Couchman 2U book 2, chapter 19.10 question 5.


How do i integrate this? (This is in the chapter of The Derivative of y=a^x)



Thanks!!

 

[tommykins]