Differentiating Logs (2 Viewers)

[Timske]

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 ,

 

[Coookies]

Where did the 1/x come from?

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

 

[Timske]

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)

 

[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.

 

[Timske]

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

 

[Coookies]

Is it because elnx = xlne?

 

[D94]

Lnx=1 get rid of ln by multiplying by e

elnx = e
x=e

You would be exponentially raising each side, ie. e^ln(x) = e¹.

 

[Coookies]

Would they expect a maths adv student to do that?
Multiplying is much less complicated lol!

 

[Coookies]

One last question, how would I test it? lol, whats on either side of e?

 

[Timske]

You would be exponentially raising each side, ie. e^ln(x) = e¹.

Oh lol, and yes u will need to know this its not hard at all

 

[D94]

Would they expect a maths adv student to do that?
Multiplying is much less complicated lol!

Yes, of course. What do you mean "multiplying"? There isn't multiplying :s To remove the logarithm, you take the exponents of both sides; that's the best method.

 

[D94]

One last question, how would I test it? lol, whats on either side of e?

e is approximately 2.718..., so you can test using say 2 and 3.

 

[Coookies]

So once I've raised them both, how do I get to x=e?

& I tried that but its all increasing (answer says max)

 

[D94]

So once I've raised them both, how do I get to x=e?

& I tried that but its all increasing (answer says max)

e^log_e(x) = e¹

Hm, it sounds like you haven't been given a good explanation yet. One of the most important concepts is to raise log by e or to raise e by log; they are sort of like inverses of each other, and can cancel each other out. But, not like multiplication/division, they are powers of each other. (You're going to need a better explanation of the whole log and e relation, because it's too easy to just take these things for granted)

Anyway, it should be a maximum; let x = 2, y' = (1 - ln(2))/4 > 0, and when x = 3, y' = (1 - ln(3))/9 < 0.