Help with Deritvitives dealing with limits

**cjmac87** · September 8th 2008, 01:55 PM

Got caught up here... would greatly appreciate some help in working through these...

An object moves along so that it's location is -5x + 2x . Find velocity at following points: 1, 3, 5.
This is using derivatives and finding the velocities at the points. I've used what i know to find the slope of the tangent and secant line... unless I did my math wrong thats not how to find the answer. Any advise please?

Also, if someone can show me a method of solving a derivitive with a fraction, that'd be great.

$Ex. f(x) = \frac {1}{x + 2}$

**11rdc11** · September 8th 2008, 02:10 PM

I'm not sure what you asking but just take the derivative of your position function to get your velocity function.

From there just plug in the points given to you into your velocity function to find it velocity.

**cjmac87** · September 8th 2008, 02:59 PM

I did exactly as you said, and the correct answer has not came up for me.

And I could really use help on solving with a fraction if possible.

**11rdc11** · September 8th 2008, 03:11 PM

$f(x) = \frac {1}{x + 2} = \frac{u}{v}$

using quotient rule

$\frac{(du)(v) - (u)(dv)}{v^2}$

$-\frac{1}{(x + 2)^2}$

**11rdc11** · September 8th 2008, 03:14 PM

For your position function

$f(x) = - 5x + 2x$

$f(x) = -3x$

now take the derivative

$f'(x) = -3$

so the velocity is -3 at any time

**cjmac87** · September 8th 2008, 03:33 PM

Sorry, bad typo... i mean [tex] -5x^2 +2x [tex]

**Krizalid** · September 8th 2008, 04:09 PM

Having $f(x)=\frac1{x+2},$ it's a worth of try on writtin' it as $f(x)=(x+2)^{-1},$ now this is a matter of power rule + chain rule.

**11rdc11** · September 8th 2008, 04:33 PM

Thanks Krizalid but I remember learning quotient rule 1st before learning chain rule so thought he may not know how to apply chain rule just yet.

$f(x) = -5x^2 +2x$

$f'(x) = -10x + 2$

Now plug in your points {1,3,5}

f'(1) = -8
f'(3) = -28
f'(5) = -48

Just wondering you said your doing derivatives using limits so do you want to see the the derivative found like this?

$\lim_{h \to 0} \frac{[-5(x + h)^2 + 2(x + h)] -[ -5x^2 +2x ]}{h}$

**cjmac87** · September 8th 2008, 06:41 PM

That did the trick... although that's what I seemed to be doing the entire time. However... it appears i got the (-) mixed up while distributing.

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