# First Principles Calculus

• Jun 6th 2008, 11:01 AM
First Principles Calculus
Ok i got this homework sheet out.
There are a LOT of questions to do.

I have had one lesson on calculus 1st principles and I'm a bit confused.

I've attached a file of the sheet. For the second question would I use first principles? It doesn't say and that is a LOT of working for the first question 2a) I was nearly over two pages.

(Yeh I'm new to Advanced Higher course in scotland!)

http://www.mathhelpforum.com/math-he...attach/jpg.gif
• Jun 6th 2008, 11:04 AM
janvdl
Quote:

Ok i got this homework sheet out.
There are a LOT of questions to do.

I have had one lesson on calculus 1st principles and I'm a bit confused.

I've attached a file of the sheet. For the second question would I use first principles? It doesn't say and that is a LOT of working for the first question 2a) I was nearly over two pages.

(Yeh I'm new to Advanced Higher course in scotland!)

http://www.mathhelpforum.com/math-he...attach/jpg.gif

With all due respect, you must be dreaming if you think we are going to do your homework for you.

It would appear you do not have to use first principles for question 2.
• Jun 6th 2008, 11:06 AM
Also for question 1d) and the like I struggle to understand them.
Here's my working so far:
lim h->0 = 1/(x+h) - 1/x
= x/x(x+h) - (x+h)/x(x+h)
= x - (x+h)/x(x+h)
=h/x(x+h)

???

Am I right? Hard to type the answer. I could scan my working to show you.
Oh and please i don't want people to do my homework for me! I just need a little guidance with question wording and what they're asking for.
• Jun 6th 2008, 11:07 AM
Mathstud28
Quote:

Ok i got this homework sheet out.
There are a LOT of questions to do.

I have had one lesson on calculus 1st principles and I'm a bit confused.

I've attached a file of the sheet. For the second question would I use first principles? It doesn't say and that is a LOT of working for the first question 2a) I was nearly over two pages.

(Yeh I'm new to Advanced Higher course in scotland!)

http://www.mathhelpforum.com/math-he...attach/jpg.gif

I assume that you mean $f'(x)=\lim_{\Delta{x}\to{0}}\frac{f(x+\Delta{x})-f(x)}{\Delta{x}}$

And secondly I would suggest emailing your instructor, for there is no way we could no

But if you have only been taught it this way, then I would assume so
• Jun 6th 2008, 11:10 AM
aye that's right!
Except we use an h instead of the whole triangled x
• Jun 6th 2008, 11:11 AM
Mathstud28
Quote:

Also for question 1d) and the like I struggle to understand them.
Here's my working so far:
lim h->0 = 1/(x+h) - 1/x
= x/x(x+h) - (x+h)/x(x+h)
= x - (x+h)/x(x+h)
=h/x(x+h)

???

Am I right? Hard to type the answer. I could scan my working to show you.
Oh and please i don't want people to do my homework for me! I just need a little guidance with question wording and what they're asking for.

$\lim_{\Delta{x}\to{0}}\frac{\frac{1}{x+\Delta{x}}-\frac{1}{x}}{\Delta{x}}=\lim_{\Delta{x}\to{0}}\fra c{\frac{-\Delta{x}}{x^2+x(\Delta{x})}}{\Delta{x}}=\lim_{\De lta{x}\to{0}}\frac{-1}{x^2+x(\Delta{x})}=\frac{-1}{x^2}$
• Jun 6th 2008, 11:15 AM
galactus
For first principles, plug in x+h for x for the f(x+h) portion. Like so:

#1: $\lim_{h\rightarrow{0}}\frac{\overbrace{3(x+h)^2+2( x+h)}^{\text{f(x+h)}}-(\overbrace{3x^{2}+2x}^{\text{f(x)}})}{h}$

Simplify and cancel. Then you should see the derivative. Which is obviously

$6x+2$
• Jun 6th 2008, 11:16 AM
Quote:

Originally Posted by galactus
For first principles, plug in x+h for x for the f(x+h) portion. Like so:

#1: $\lim_{h\rightarrow{0}}\frac{\overbrace{3(x+h)^2+2( x+h)}^{\text{f(x+h)}}-(\overbrace{3x^{2}+2x}^{\text{f(x)}})}{h}$

Simplify and cancel. Then you should see the derivative. Which is obviously

$6x+2$

hmmm yes but that's a lot of working and I've got lots more questions over the page...
I'm unsure if I'm being expected to 1st princ. it all
Coz its a lot of simplifying
• Jun 6th 2008, 11:18 AM
galactus
It looks more ominous than it is. Give it a go. You'll see it as h-->0.
• Jun 6th 2008, 11:28 AM
Ah you're right once you've expanded and collected together h's
it is more simple!

Would I be expected to do the same with fractional powers??
• Jun 6th 2008, 11:32 AM
Oh wait I'm seeing now tooo!!
in Q2c)
if you add the powers 4/3 and 1/3 you'll get 5/3

and you can take away 4/3
to give 1/3

... change into a cube rooot?
• Jun 6th 2008, 11:32 AM
Mathstud28
Quote:

Ah you're right once you've expanded and collected together h's
it is more simple!

Would I be expected to do the same with fractional powers??

Fractional powers cannot be expanded at the level of beginning calculus, so no. You would not be expected to do it that way
• Jun 6th 2008, 12:25 PM
galactus
Let's do a fractional power. It's a little trickier in that you can use the conjugate.

Take $x^{\frac{3}{2}}$. Let's find the derivative through first principles.

$\frac{(x+h)^{\frac{3}{2}}-x^{\frac{3}{2}}}{h}\cdot\frac{(x+h)^{\frac{3}{2}}+ x^{\frac{3}{2}}}{(x+h)^{\frac{3}{2}}+x^{\frac{3}{2 }}}$

Multiply through and simplifying:

$\lim_{h\to\rightarrow{0}}\frac{3hx^{2}+3xh^{2}+h^{ 3}}{h((x+h)^{\frac{3}{2}}+x^{\frac{3}{2}})}$

Now, as we can see by h-->0, we have :

$\lim_{h\to\rightarrow{0}}\frac{3x^{2}}{(x+h)^{\fra c{3}{2}}+x^{\frac{3}{2}}}+\lim_{h\to\rightarrow{0} }\frac{3xh}{(x+h)^{\frac{3}{2}}+x^{\frac{3}{2}}}+\ lim_{h\to\rightarrow{0}}\frac{h^{2}}{(x+h)^{\frac{ 3}{2}}+x^{\frac{3}{2}}}$

Which gives us:

$\frac{3x^{2}}{2x^{\frac{3}{2}}}=\boxed{\frac{3\sqr t{x}}{2}}$

Which is as it should be.(Clapping)

Does that help?. A wee bit anyway?.
• Jun 7th 2008, 06:40 AM
xD Cheers yeh it will help.
Shall work through that tonight I can't take it in right now *fried brain*

Thanks for all the help
• Oct 12th 2008, 04:36 AM
Andrew1380
Urgent Help - Differentiation
Hi

Am totally lost, leaning from a book as I need to start my semester in finance next year Jan, am preparing for it, Please guide?

c=0.004q(to the power3)+20q+5000

and the demand function is p=450-4q

Thanks(Worried)

Find the profit-maximizing output?