derivatives problems

nbalakerskb · 2006-09-17 08:06:22

f(x)= 1/ square root of x+2
find the f '(a)

lim h->0 (cos( 3.14 +h) + 1)/h
find the f (x) and a.
thx alot

mrsesyc · 2006-09-17 09:16:52

well you need to change the form to get it in the derivative form - so f'(a) = (x+2) to the power of ½ so you multiply the ½ infront of (x+2) and substract 1 from the power which would be ½(x+2) and when u substract one from the power u get -½ and do the derivative of x would be 1 so f' = ½(x+2) to the -½ as the power *1 . how ever i am not sure what your limit is trying to say if u can resend the limit part clearly

mahmoudaljamel · 2006-09-17 19:45:40

f(x) = 1/ ( x+2)½
= ( x+ 2)- ½
f'(x)= [-1/2 ( x+2 )- ³/ ² ]* 1
= -1 / [ 2 (x+2)³/ ²]
f'(a)= -1 / [2 (a+2)³/ ² ]
f'(2)= -1 / [2 (4)³/ ²]
= -1 / 16

can you please rewrite the second question

Last edited by mahmoudaljamel (2006-09-17 20:00:51)

John E. Franklin · 2006-09-18 21:26:16

Hi nbalak, for the second question: lim h->0 (cos( 3.14 +h) + 1)/h
I have a possible answer, which is just a little less than infinity, so pretty much infinity.
The answer I get is: infinity - 1/(two infinities)
So that is infinity minus the reciprocal of twice infinity, that last part being nearly zero, so close to infinity.
I did it in radians. The sine of a small radian is close to the small radian.
Then I used pythagoreans theorm to get cosine or x-shadow.
x-shadow length = sqroot(1 - smallRadianSquared)
I used .001 for my testing and noticed that .99999949 was coming out, so that was really negative at 3.14159 angle (180 degrees).
I added the 1 you have in equation.
Reminds me of haversine, can't remember.
Hope this shabby prose helps a bit...

John E. Franklin · 2006-09-18 21:34:04

quote from wikipedia:
The versed sine, also called the versine and, in Latin, the sinus versus ("flipped sine") or the sagitta ("arrow"), is a trigonometric function versin(θ) (sometimes further abbreviated "vers") defined by the equation:

John E. Franklin · 2006-09-18 21:58:10

Nice picture from Wiki:
Image:Versin.png DOESN'T DISPLAY FOR ME??

http://en.wikipedia.org/wiki/Image:Versin.png THIS ONE CLICKS AND WORKS!!

Image:Versin.png DOESN'T DISPLAY FOR ME??

Last edited by John E. Franklin (2006-09-18 22:02:33)

Dross · 2006-09-19 01:26:10

nbalakerskb wrote:

lim h->0 (cos( 3.14 +h) + 1)/h

John, I don't know how you got infinity... ???

I get an answer of 0 for just this (ignoring the bit afterwards, which is mildly confusing!).

Using L'Hopital's rule, since

, we get that:

John E. Franklin · 2006-09-19 15:32:13

I don't remember L'Hopital's Rule, but maybe I'll relearn it someday and get back to you.
Why do you replace the h in denominator with a one, when the limit is h goes to zero??
I'm not positive my answer is correct, but I could look into it further if needed.

Dross · 2006-09-19 21:42:06

The h in the denominator is replaced by a 1 because L'Hopital's rule essentially states that:

L'Hopital's rule also works for

and .

John E. Franklin · 2006-09-20 02:47:43

Thanks for explaining L'Hopital's, I just also read elsewhere as well.
Your work looks great, so then I rechecked my work, this time actually using paper instead of everything in my head and a calculator in hand. And it turns out for small h's near zero, the answer is one-half of the inputted near-zero number.
So my original mistake did have the 1 over twice infinity, but I went hay-wire somewhere.
For 1/1000, you get about 1/2000
For 1/18181.8, you get about 1/36351.6, using the pythagorean theorm way.
[squareRoot(1-h up 2) - 1 ] / h
where h up 2 means h^2 or h squared.
I forgot some Latex, so sorry. Thanks for the L'Hopital lesson!!

Math Is Fun Forum

#1 2006-09-17 08:06:22

derivatives problems

#2 2006-09-17 09:16:52

Re: derivatives problems

#3 2006-09-17 19:45:40

Re: derivatives problems

#4 2006-09-18 21:26:16

Re: derivatives problems

#5 2006-09-18 21:34:04

Re: derivatives problems

#6 2006-09-18 21:58:10

Re: derivatives problems

#7 2006-09-19 01:26:10

Re: derivatives problems

#8 2006-09-19 15:32:13

Re: derivatives problems

#9 2006-09-19 21:42:06

Re: derivatives problems

#10 2006-09-20 02:47:43

Re: derivatives problems

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