Anyone good at Calculus ?

I have a calculus assignment due in next week, and needed some help on a few questions. They are easy, though since I have been off school for 2 months I have forgotten lots so I don't remember how to do them.

If anyone can do these Q's, it would be great . Maybe if your bored try them out. Thanks.


link: http://img151.imageshack.us/img151/6471/calca1av6.jpg


link: http://img151.imageshack.us/img151/5803/calca2or5.jpg


link: http://img260.imageshack.us/img260/346/calca3fo4.jpg

- Phatmat.

 

I cant see any questions. and ur siggy is distracting me.

 

The questions show up for me, I don't see why they wouldn't work. I will try to post up the links to the pictures.

Lol about the sig .

- Phatmat.

 

I don't have any time this weekend, but if no one else does them I will try to do them next week.

 

ironically, i have no idea what that says.


my kind of math is more like 2+4 =p

 

I don't think anyone could answer on Sythe. Try asking at yahoo answers or something.

I'm doing calculus next year (maybe) so if you need help then, ask me lol =P

 

Thanks Steph. The assignment is due in this Thursday, so if you get a chance to do any it would be good.

If you don't have the time it should be ok.

- Phatmat.

 

I'm working on a few of them at the moment. I'll edit in some solutions in a little while.

I started a math help/discussion thread a little while ago. Feel free to ask any more questions there if you need: http://sythe.org/showthread.php?t=369987.

EDITED IN:
If you don't understand any of the steps I did, I would be glad to expound on anything. I have been doing calculus for a little while, so I might be skipping some steps that you aren't as familiar with.

2. (e)

Code:

   y = (e^x)ln(2x)

Use product rule:

y' = (e^x)' * ln(2x) + (e^x) * (ln(2x))'

= (e^x) * ln(2x) + (e^x) * 1/(2x)*2    <--- use chain rule to find (ln(2x))'

Now just a little factoring and simplifying:

= (e^x)(ln(2x) + 1/x)  done.

3. (a)

Code:

 Ok, we are trying to prove this: d/dx [f(x)]^-1 =  -f'(x)/[f(x)]^2.

If we simply do the derivative on the left side of the equation, we will get the stuff on the right.

d/dx [f(x)]^-1 = -[f(x)]^-2 [b] * f'(x)[/b] <--- using chain rule

Now just simplify:

=  -f'(x)/[f(x)]^2. <-- this is what we were trying to get.

3. (b)

Code:

 f(x) * [g(x)]^-1

Use product rule:

f'(x) * [g(x)]^-1 + f(x) * ([g(x)]^-1)'

Now we need to find: ([g(x)]^-1)'. Hey, look! We did that in the previous part of the question. So, I think I'll just put it in instead of doing it again.

= f'(x) * [g(x)]^-1 + f(x) * -g'(x)/[g(x)]^2

and simplify:

= ( f'(x) / g(x) ) - (f(x)*g'(x)) / [g(x)]^2

This is indeed the answer although it is not in the same form as the quotient rule is usually written. To get that form,  multiply  ( f'(x) / g(x) ) by g(x)/g(x). We can do this since g(x)/g(x) is simply equal to 1.

= (f'(x) * g(x)) / [g(x)]^2 - (f(x)*g'(x)) / [g(x)]^2

Now both parts of this have the same denominator. So, subtract:

[IMG]http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/quotientruledirectory/img2.gif[/IMG]

I realize the notation is hard to read... Tell me if any part doesn't make sense.

Anyway, I really don't feel like doing the others at the moment. Why don't you try them, and if any of them don't make sense, come back and ask. Good luck!

 

www.mathhelpforum.com

No, really.

 

Thanks greatly isosceles, if there was a +rep system I would have given you a lot.

I am just reading 3. (b) now, and trying to interpret the notation, but I can understand your steps. Thanks.

- Phatmat.