hihi T_T

New Problem:X^3+Y^3=125

dy/dx =?
a. -x/y
b.-x/5y
c. y/x
d. x/(y-5)
e. x/y

Well, I thought it was -(x/y)^2, but it's not on there.
I think that it should be negative, and there should not be a 5, which leaves me -x/y, which is x^2+y^2=C

Problem: The function y=|x-8|+2 is differentiable over..

a. [8,10)
b. (2,10)
c. [2,10]
d. [2,10)
e. (2,10]

halps would be appreciated <3

Edit: What we got is that since [8,10) has a continuous slope of 1, it's right.

Saracen

United States5139 Posts

October 01 2008 04:11 GMT

graph it
if the sharp point from the absolute value is included in your set of points, then no
if not, then yes

doghunter

United States23 Posts

October 01 2008 04:13 GMT

It's not differentiable at x=8 (cusp), so I thought the answer would be (-inf,8),(8,inf) but that doesn't seem to be an answer.

Avidkeystamper

United States8551 Posts

October 01 2008 04:13 GMT

It's a, all the other intervals include the corner of the V the graph produces and that corner is not differentiable since the slope from both sides is not the same. Notice how a does not include 8 in its interval since that point is part of the corner and is not differentiable (<-I think?)

FragKrag

United States11262 Posts

October 01 2008 04:18 GMT

We did graph it, and it's a V, but 2,10 all contain two slopes, 1 and -1 as derivatives.

doghunter

United States23 Posts

October 01 2008 04:32 GMT

There's no limit from the left side, so you can't say that there is a limit at x=8

vAltyR

United States581 Posts

October 01 2008 04:36 GMT

what it's asking is if you get a continuous derivative over the intervals. If I'm interpreting the equation right, the point of the "V" should be at x=8, therefore the derivative at x=8 does not exist, eliminating all the intervals between 2 and 10, whether or not they actually *include* 2 or 10. the answer is a.

man, that's a *precalc* question? we never covered that until calculus. in fact, i never even heard the word "derivative" until i opened a calc book.

thunk

United States6233 Posts

October 01 2008 04:48 GMT

8 is a point of non-differentiability, so I think that (a) is also wrong. But for sure b-e are wrong more so, so it's definitely A.

On October 01 2008 13:32 doghunter wrote:
There's no limit from the left side, so you can't say that there is a limit at x=8

It's continuous and the limit as x approaches 8 exists. It's just no differentiable because the derivative from the right and the derivative from the left are inequal.

Purind

Canada3558 Posts

October 01 2008 14:56 GMT

See I thought a would be wrong too, but if you only graph that interval, you get a straight line and you don't see the pointiness

SIUnit

China288 Posts

October 01 2008 16:29 GMT

#10

Asia tend to be a bit crazy on the math side of things. And yeah, this would technically be a calc 1 question here in the states.

FragKrag

United States11262 Posts

October 22 2008 03:28 GMT

#11

bump for new question~

skyglow1

New Zealand3962 Posts

October 22 2008 04:46 GMT

#12

Use implicit differentiation which involves chain rule.
x^3+y^3=125
Differentiation both sides by x: 3x^2+d(y^3)/dx=0
Since y is a function of x, and using chain rule: d(y^3)/dx = 3y^2 x dy/dx
So you get: 3x^2 + 3y^2 x dy/dx = 0
And rearranging: dy/dx = -3x^2/3y^2

[Edit] I'm a retard -_-

SaveYourSavior

United States1071 Posts

October 22 2008 08:18 GMT

#13

what i got was -x^2 / y^2

but

a seems to be the right answer since it seems the most right

Divinek

Canada4043 Posts

October 22 2008 08:30 GMT

#14

On October 22 2008 13:46 skyglow1 wrote:
Use implicit differentiation which involves chain rule.
x^3+y^3=125
Differentiation both sides by x: 3x^2+d(y^3)/dx=0
Since y is a function of x, and using chain rule: d(y^3)/dx = 3y^2 x dy/dx
So you get: 3x^2 + 3y^2 x dy/dx = 0
And rearranging: dy/dx = -3x^2/3y^2 = -x/y so the answer should be a)

Implicit differentiation is in pre calc?

Ecael

United States6503 Posts

October 22 2008 09:10 GMT

#15

On October 22 2008 17:30 Divinek wrote:

Show nested quote +

Implicit differentiation is in pre calc?

If memory serves schools in Taiwan covered power rule to implicit differentiation in precalc, at least that's what I recall of what my friends/parents/family's talks of how they had it tough :p

EDIT - It'd be hilarious when OP posts to say how he isn't at Taiwan though.

FragKrag

United States11262 Posts

October 22 2008 14:56 GMT

#16

I thought it was a nationality? T__________T

I'm taking "PreCalculus". It's the course name, but the book has PreCalc and Calc I so lol tt;;

anyways, how can you go from -(x/y)^2 to -x/y? ?:/

skyglow1

New Zealand3962 Posts

October 22 2008 15:41 GMT

#17

Sorry I'm so retarded. I shouldn't try to help at math when I don't learn it anymore

FragKrag

United States11262 Posts

October 23 2008 00:49 GMT

#18

;_________;

Today my math teacher tells me that "abcde" is a choice when you cant find the answer. --;

onihunter

United States502 Posts

October 23 2008 01:23 GMT

#19

On October 23 2008 09:49 FragKrag wrote:
;_________;

Today my math teacher tells me that "abcde" is a choice when you cant find the answer. --;

Nice, would explain a lot about why I couldn't figure this problem out lol. And man I hate it when math questions have "None of the above" as a choice.