Reply With QuoteQuestion 1:
Question 2:
Question 3:
}{x-2} \right ) \\\\ = \frac{\frac{1}{x} \times (x-2) - \ln(x)}{(x-2)^2} \qquad $ by the Quotient Rule$ \\\\\ = \frac{\frac{x-2}{x} - \ln(x)}{(x-2)^2} \\\\ = \frac{x-2 - x\ln(x)}{x(x-2)^2})
r) x^3loge(x+1)
s) loge(logex)
t) lnx/x-2
Thanks!!!
Question 1:
Question 2:
Question 3:
Bachelor of Science (Adv. Mathematics) - University of Sydney:
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
~, ~ 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)) "\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))")
Last edited by Timske; 30 Apr 2012 at 11:32 PM.
B ECO@USYD 2013
Oh for some reason, I read the inside of the log function for Q1 to be x instead of x+1.
 \right ) \\\\ = 3x^2 \ln(x+1) + x^3 \times \frac{1}{x+1} \qquad $ by the Product Rule$ \\\\ = 3x^2 \ln(x+1) + \frac{x^3}{x+1} \\\\ = x^2 \left [ 3 \ln(x+1) + \frac{x}{x+1} \right ] )
Last edited by Carrotsticks; 30 Apr 2012 at 11:31 PM.
Bachelor of Science (Adv. Mathematics) - University of Sydney:
Excuse my hideous latex
B ECO@USYD 2013
For Q2, could you write the derivative of loge?
Originally Posted by Coookies
Bachelor of Science (Adv. Mathematics) - University of Sydney:
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.
Bachelor of Science (Adv. Mathematics) - University of Sydney:
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?
Bachelor of Science (Adv. Mathematics) - University of Sydney:
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:
} \times \frac{1}{x} = \frac{1}{x \ln(x)})
Bachelor of Science (Adv. Mathematics) - University of Sydney:
ln(ln (x)) is same as loge(loge(x))
Last edited by Drongoski; 1 May 2012 at 12:15 AM.
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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.
Bachelor of Science (Adv. Mathematics) - University of Sydney:
I think thats what confuses me haha!
Thanks a lot though!
Id rep but I have to spread more first... lol
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
Last edited by Coookies; 3 May 2012 at 12:07 AM.
Do you have trouble differentiating* log?
B ECO@USYD 2013
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
d/dx lnx - x^2 + 1 - x = -2x + 1/x - 1 ,
B ECO@USYD 2013
Where did the 1/x come from?
Also, isn't ln for loge? Just wondering...
loge = ln
d/dx ln(fx) = f'x/fx
hence, d/dx lnx = 1/x
logx-x^2 + 1 - x
d/dx logx = 1/x
d/dx -x^2 = -2x
d/dx 1 = 0
d/dx -x = -1
:. -2x + 1/x -1
Log(x) = ln(x) = loge(x)
Last edited by Timske; 3 May 2012 at 12:35 AM.
B ECO@USYD 2013
Thanks!
I saw logx-x^2 as a whole thing. Didn't know you could separate them.
For Q10, is the differentiation (1-lnx)/x^2 right?
If so, when I make it equal 0, I get lnx=1 and then I don't know what to do.
Lnx=1 get rid of ln by multiplying by e
elnx = e
x=e
B ECO@USYD 2013
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