C3 January 2013 25/01/2013

Can differentiate cot(x) ... by making it equal to cos(x)/sin(x)

How about if I make it equal [tan(x)]^-1
and I differentiate that...

Also how would I differentiate cosec and sec ?? :smile:

Thanks in advance

To differentiate cosec(x), write it as 1/sin(x) and use the quotient rule.
To differentiate sec(x), write it as 1/cos(x) and use the quotient rule.

:smile:

Reply 1841

Commenting94

Original post by Alex_1

What kind of questions is everyone worried about the most?
Any last minute tips?

Posted from TSR Mobile

Worried about any simultaneous equations on exponentials and logs. They seem to catch me out with silly mistakes all the time.

Although its hardly come up, there might be a chance it will today.

Maybe a question involving half-angles will appear today?

Reply 1842

lewiss111

Original post by ZujajahM

To differentiate cosec(x), write it as 1/sin(x) and use the quotient rule.
To differentiate sec(x), write it as 1/cos(x) and use the quotient rule.

:smile:

its the chain rule, as its not two functions
because sec(x) = (cos(x))^-1 = sin(x)(cos)^-2 = sec(x)tan(x)

Reply 1843

`Mo

Original post by lewiss111

its the chain rule, as its not two functions
because sec(x) = (cos(x))^-1 = sin(x)(cos)^-2 = sec(x)tan(x)

Either way works, there's no difference.

Reply 1844

MedMed12

21

Original post by Commenting94

Worried about any simultaneous equations on exponentials and logs. They seem to catch me out with silly mistakes all the time.

Although its hardly come up, there might be a chance it will today.

Maybe a question involving half-angles will appear today?

simultaneous equations? :O

Reply 1845

Corvus

I have a quick question, does y= ln(x)+3 has a asymptote at y=3? Thanks

Reply 1846

posthumus

20

Original post by Commenting94

Worried about any simultaneous equations on exponentials and logs. They seem to catch me out with silly mistakes all the time.

Although its hardly come up, there might be a chance it will today.

Maybe a question involving half-angles will appear today?

What sorts do you get stuck on ?

Like when you end up with something like e^2ln(x)

?? or have you come across harder ones :smile:

Reply 1847

Commenting94

Simultaneous equations like this one I came across yesterday:

Solve:
e^2y - x + 2 = 0
ln(x+3) - 2y - 1 = 0

Reply 1848

MedMed12

21

Original post by Commenting94

Simultaneous equations like this one I came across yesterday:

Solve:
e^2y - x + 2 = 0
ln(x+3) - 2y - 1 = 0

which paper? :O

Reply 1849

TOSully

Original post by Corvus

I have a quick question, does y= ln(x)+3 has a asymptote at y=3? Thanks

Yes, the asymptote is at y = 3 because it has shifted up 3 units in a positive direction on the y-axis. This means that the former asymptote of y = 0 has also shifted up 3 units, therefore, asymptote at y = 3. Hope this helps :smile:

Reply 1850

Meme4

1

Original post by TOSully

Yes, the asymptote is at y = 3 because it has shifted up 3 units in a positive direction on the y-axis. This means that the former asymptote of y = 0 has also shifted up 3 units, therefore, asymptote at y = 3. Hope this helps :smile:

Y=ln x has an asymptote x=0

Reply 1851

Commenting94

Original post by MedMed12

which paper? :O

Really sorry I can't rememeber which paper, I done a couple random practice papers and solomons yesterday, It was from one of them.

I don't think its appeared in any of the past papers from 06-12 though. Just a small possibilty it might come up today?

Reply 1852

Henry.Lister

- How do i do part c?

I tried linking the first differentiation with 0 but i don't get the right answer!

(edited 11 years ago)

Original post by Commenting94

Simultaneous equations like this one I came across yesterday:

Solve:
e^2y - x + 2 = 0
ln(x+3) - 2y - 1 = 0

ln(5+e^2y) - 2y -1 = 0

hmm so what happens now ?

5+e^2y - e^2y- e = 1

4-e=0 :tongue:

HELP !!

Reply 1854

steve2005

17

Original post by Skaterkid

all of that equals to 1

Why do you think (tan(22.5))^2 equals 1 ?

Reply 1855

Meme4

1

Original post by Corvus

I have a quick question, does y= ln(x)+3 has a asymptote at y=3? Thanks

No. Y=ln(x) has an asymptote at x=0
+3 moves it up 3
Asymptote is still at x=0

Reply 1856

This Excellency

Original post by Commenting94

Simultaneous
equations like this one I came across yesterday:
Solve:
e^2y - x + 2 = 0
ln(x+3) - 2y - 1 = 0

1) e^{2y} - x + 2 = 0
2) ln(x+3) - 2y - 1 = 0

y=\frac{1}{2}ln(x-2)

Sub into 2)

ln(x+3)-ln(x-2)-1=0

ln \frac{(x+3)}{x-2}=1

e^1=\frac{x+3}{x-2}

Solve from there :smile:

Reply 1857

Commenting94

Original post by This Excellency

1) e^{2y} - x + 2 = 0
2) ln(x+3) - 2y - 1 = 0

y=\frac{1}{2}ln(x-2)

Sub into 2)

ln(x+3)-ln(x-2)-1=0

ln \frac{(x+3)}{x-2}=1

e^1=\frac{x+3}{x-2}

Solve from there :smile:

Yep exactly (:

I just made silly mistakes, and panicked. But got it at the end :smile:

Reply 1858

TOSully

Original post by Meme4

Y=ln x has an asymptote x=0

I completely misread the question lol. I read it quickly and thought we were talking about e^x. Yes, y = ln(x) has an asymptote x = 0, not y = 0. Therefore, ln(x)+3 doesn't affect the x = 0 asymptote but wil change the coordinate of the x-axis intercept from (1,0) to (e^-3,0).

Reply 1859

Meme4

1

Original post by TOSully

I completely misread the question lol. I read it quickly and thought we were talking about e^x. Yes, y = ln(x) has an asymptote x = 0, not y = 0. Therefore, ln(x)+3 doesn't affect the x = 0 asymptote but wil change the coordinate of the x-axis intercept from (1,0) to (e^-3,0).

Thank god for that. I thought I had got something badly wrong :tongue:

C3 January 2013 25/01/2013

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