1. ## Differentiating Logs

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

Thanks!!!

2. ## Re: Differentiating Logs

Question 1:

$\\ \frac{d}{dx} \left ( x^3 \ln(x) \right ) \\\\ = 3x^2 \ln(x) + x^3 \times \frac{1}{x} \qquad by the Product Rule \\\\ = 3x^2 \ln(x) + x^2 \\\\ = x^2 \left ( 3 \ln(x) + 1 \right )$

Question 2:

$\\ \frac{d}{dx} \left ( \ln \left ( \ln(x) \right ) \right ) \\\\ = \frac{1}{\ln(x)} \times \frac{1}{x} \qquad by the Chain Rule \\\\ = \frac{1}{x \ln(x)}$

Question 3:

$\\ \frac{d}{dx} \left ( \frac{\ln(x)}{x-2} \right ) \\\\ = \frac{\frac{1}{x} \times (x-2) - \ln(x)}{(x-2)^2} \qquad by the Quotient Rule \\\\\ = \frac{\frac{x-2}{x} - \ln(x)}{(x-2)^2} \\\\ = \frac{x-2 - x\ln(x)}{x(x-2)^2}$

4. ## Re: Differentiating Logs

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

$\\ \frac{d}{dx} \left ( x^3 \ln(x+1) \right ) \\\\ = 3x^2 \ln(x+1) + x^3 \times \frac{1}{x+1} \qquad by the Product Rule \\\\ = 3x^2 \ln(x+1) + \frac{x^3}{x+1} \\\\ = x^2 \left [ 3 \ln(x+1) + \frac{x}{x+1} \right ]$

5. ## Re: Differentiating Logs

Excuse my hideous latex

6. ## Re: Differentiating Logs

For Q2, could you write the derivative of loge?

7. ## Re: Differentiating Logs

Originally Posted by Coookies
For Q2, could you write the derivative of loge?
$\frac{d}{dx} \left [ \log_e x \right ] = \frac{1}{x}$

8. ## Re: Differentiating Logs

Originally Posted by Carrotsticks
$\frac{d}{dx} \left [ \log_e x \right ] = \frac{1}{x}$
Theres no x though

9. ## Re: Differentiating Logs

Originally Posted by Coookies
Theres no x though
Oh, so you just wanted the derivative of the constant:

$\log_e e$

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

10. ## Re: Differentiating Logs

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

11. ## Re: Differentiating Logs

Originally Posted by Coookies
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?

12. ## Re: Differentiating Logs

This bit:

s) loge(logex)

haha..

13. ## Re: Differentiating Logs

Originally Posted by Coookies
This bit:

s) loge(logex)

haha..
We use the Chain Rule.

The derivative of:

$\ln \left ( f(x) \right )$

is:

$\frac{f'(x)}{f(x)}$

by the Chain Rule.

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

$\frac{1}{\ln(x)} \times \frac{1}{x} = \frac{1}{x \ln(x)}$

14. ## Re: Differentiating Logs

ln(ln (x)) is same as loge(loge(x))

15. ## Re: Differentiating Logs

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...

16. ## Re: Differentiating Logs

Originally Posted by Coookies
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.

17. ## Re: Differentiating Logs

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

18. ## Re: Differentiating Logs

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

19. ## Re: Differentiating Logs

Originally Posted by 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
Do you have trouble differentiating* log?

20. ## Re: Differentiating Logs

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

21. ## Re: Differentiating Logs

Originally Posted by 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
d/dx lnx - x^2 + 1 - x = -2x + 1/x - 1 ,

22. ## Re: Differentiating Logs

Where did the 1/x come from?

Also, isn't ln for loge? Just wondering...

23. ## Re: Differentiating Logs

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)

24. ## Re: Differentiating Logs

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.

25. ## Re: Differentiating Logs

Originally Posted by Coookies
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

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