Comments | sariebeth88: Another Calculus Question

sariebeth88

Another Calculus Question

Apr 22, 2008 00:09

So I'm doing the integrals of the inverse derivatives...blah
HELP PLEASE!I've done the whole problem I'm just stuck on the easy part so any help would be amazing ( Read more... )

Comments 9

(The comment has been removed)

sariebeth88 April 22 2008, 04:24:27 UTC

she said "Using L'Hopital's Rule, find" :

lim
4-4e^x
x e^x

as x-->0

mananath April 22 2008, 04:35:49 UTC

agggh my head. it hurts.

sariebeth88 April 22 2008, 04:38:48 UTC

and what have you done? does it hurt just because you read it?

haha I HAVE TO DO IT! I tortured myself by taking Calc II and now I have to look at this everyday!

kello2008 April 22 2008, 18:35:00 UTC

i can help with L'Hospitals!

kello2008 April 22 2008, 18:37:58 UTC

lim
4-4e^x
x e^x

as x-->0

ok so to use LH it has to be 0/0...
when you plug in 0 on top you get 4-4e^(0) = 4 -4*1=0
when you plug in 0 on bottom you get 0*e^0=0*1=0
therefore, using LH,
you get...
-4e^x
e^x

= -4e^(0)
e^0

= -4/1= -4 ... YAY!

sariebeth88 April 22 2008, 20:56:29 UTC

You are my miracle worker!

I have 2 more.

LHR again...
Tell if (10x^3 + 2x^2) grows faster, slower or stays the same as e^x

Evaluate: dy/dx = x^2 y^1/2

Oh and the easy part I can't think of is what is the inverse tangent of 2?

kello2008 April 22 2008, 22:48:19 UTC

okk hm let me think!

im honestly not sure how to do the first one...
i think the second one is using differential equations?
so you get..
dy/dx= x^2 y^1/2
1/y^(1/2)=x^(2)
take the integral of both sides
you gett..
.5y^(.5)=x^(3)/3 + C
solve this for y...
y=(2x^(3)/3)^2
y= 4/9 * x^5

and the inverse tangent of 2 is... hmm.. idk! lol, i dontknow my trig well

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