|
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.
|
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 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.
|
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.
|
|
|
|