
hihi T_T
New Problem:X^3+Y^3=125
dy/dx =? a. x/y b.x/5y c. y/x d. x/(y5) 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=x8+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.

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

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.

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?)

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

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

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.

8 is a point of nondifferentiability, so I think that (a) is also wrong. But for sure be 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.

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

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.


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 _

what i got was x^2 / y^2
but
a seems to be the right answer since it seems the most right

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?

On October 22 2008 17:30 Divinek wrote:Show nested quote +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? 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.

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? ?:/

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

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

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.

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. ; Ah good, I was going a bit insane and trying to log this through to solve earlier.



