Differentiating Logs (2 Viewers)

[Coookies]

r) x^3loge(x+1)
s) loge(logex)
t) lnx/x-2

Thanks!!!

 

[Carrotsticks]

Question 1:



Question 2:



Question 3:

 

[Timske]

r) x^3Ln(x+1) use product rule , uv' + vu' = x^3/x+1 + 3x^2Ln(x+1)
= x^2[ x/x+1 + 3ln(x+1) ]

derivative of ln(x+1) = 1/x+1

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{d}{dx} ~ x^3\log_{e}(x@plus;1)~, ~ use ~product ~rule ~uv'~ @plus; ~vu'\\\\ = \frac{x^3}{x@plus;1} @plus; 3x^2\log_{e}(x@plus;1) \\\\ Factor ~ x^2 \\\\ \therefore x^2~(\frac{x}{x@plus;1} ~@plus; ~3\log_{e}(x@plus;1))" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{d}{dx} ~ x^3\log_{e}(x+1)~, ~ use ~product ~rule ~uv'~ + ~vu'\\\\ = \frac{x^3}{x+1} + 3x^2\log_{e}(x+1) \\\\ Factor ~ x^2 \\\\ \therefore x^2~(\frac{x}{x+1} ~+ ~3\log_{e}(x+1))" title="\frac{d}{dx} ~ x^3\log_{e}(x+1)~, ~ use ~product ~rule ~uv'~ + ~vu'\\\\ = \frac{x^3}{x+1} + 3x^2\log_{e}(x+1) \\\\ Factor ~ x^2 \\\\ \therefore x^2~(\frac{x}{x+1} ~+ ~3\log_{e}(x+1))" /></a>

 

[Carrotsticks]

Oh for some reason, I read the inside of the log function for Q1 to be x instead of x+1.

 

[Timske]

Excuse my hideous latex

 

[Coookies]

For Q2, could you write the derivative of loge?

 

[Carrotsticks]

For Q2, could you write the derivative of loge?

 

[Coookies]

Theres no x though

 

[Carrotsticks]

Theres no x though

Oh, so you just wanted the derivative of the constant:



If that was the question, then it would be 0, just like the derivative of any other constant.

 

[Coookies]

You're confusing me lol!
I mean just the ln in Q2

 

[Carrotsticks]

You're confusing me lol!
I mean just the ln in Q2

Now you are confusing me too! =p

You asked "could you write the derivative of loge?"

What exactly did you mean by that?

 

[Coookies]

This bit:

s) loge(logex)

haha..

 

[Carrotsticks]

This bit:

s) loge(logex)

haha..

We use the Chain Rule.

The derivative of:



is:



by the Chain Rule.

So when we differentiate ln(ln(x)), we have:

 

[Drongoski]

ln(ln (x)) is same as log_e(log_e(x))

 

[Coookies]

I'm really sorry about this, but where is the 1 from? The 1/lnx
Because isn't the derivative of lnx, 1/x?
Am I even making any sense? I don't think I even know what I'm talking about haha!
I think I'll ask my teacher tomorrow...

 

[Carrotsticks]

I'm really sorry about this, but where is the 1 from? The 1/lnx
Because isn't the derivative of lnx, 1/x?
Am I even making any sense? I don't think I even know what I'm talking about haha!
I think I'll ask my teacher tomorrow...

Indeed, explaining things via typing isn't the best way.

Yes, the derivative of ln(x) is 1/x, but we have a ln(x) WITHIN a ln(x) (ln-ception haha), so that changes things.

 

[Coookies]

I think thats what confuses me haha!
Thanks a lot though!
Id rep but I have to spread more first... lol

 

[Coookies]

I have a few more questions lol!

9. Find the point of inflection on the curve y=xlogex-x^2
For this one, I just need help differentiating once (f'(x))

10. Find the stationary point on the curve y=lnx/x and determine its nature

 

[Timske]

I have a few more questions lol!

9. Find the point of inflection on the curve y=xlogex-x^2
For this one, I just need help differentiating once (f'(x))

10. Find the stationary point on the curve y=lnx/x and determine its nature

Do you have trouble differentiating* log?

 

[Coookies]

Yes, quite Lol

My teacher helped me with this question and she went from:
logx-x^2 + 1 - x to
logx - 2x + 1

Im just not sure where the x^2 went...
However, she still got the right answer