Recent content by domyy

Thanks! Got it! =)

Homework Statement lim 2x/(sin3x) x-> 0Homework Equations lim sinx/x = 1 x->0 The Attempt at a Solution is it correct to say the following: lim 2/3 (sinx/x) x-> 0 lim 2/3 (1) x-> 0 Answer: lim 2x/sin3x = 2/3 x-> 0 Because it's on the book: cos 2x(2)/3...

I don't know if I understand it. If I have: ∫x3/√x dx = ∫x3 - 1/2 = ∫x5/2 = (x5/2 + 1)/ 5/2 +1 = (x7/2)/7/2 + C Simplyfying: 2x7/2/7 + C Is this correct? If not, what am I doing wrong? Also, I am not sure how to proceed with ∫1/√x dx If I have ∫1 dx: = 1 + C So for ∫1/√x dx I'll...

Homework Statement ∫(x3 - 3x2 + x + 1)/√x dx The Attempt at a Solution ∫x3-1/2 - 3∫x2-1/2 + ∫x1-1/2 + ∫x1-1/2 ∫x5/2 - 3∫x3/2 + ∫x1/2 + ∫1/2 (x7/2)7/2 - (3x5/2)5/2 + (x3/2)3/2 + (x3/2)3/2 + C (2x7/2)/7 - (6x5/2)/5 + 2x3/2)/3 + (2x3/2)/3 + C Thank you so much!

Thanks for the reply :) But where does the sum sign come from? Because even if it's 1 - y' I will be still multiplying, as below: -csc2 (x-y) . (1- y')

Homework Statement 2y = cot(x-y) The Attempt at a Solution 2y' = -csc2 (x-y) . (1-y)(y') Is it correct so far? My book actually has 2y' = -csc^2 (x-y) + y' csc^2(x-y) And I don't understand where that sum comes from. Am I supposed to apply the product rule? Because I tried and it didn't...

nr. pi = 3.14, right? Good you clarified that. Now I won't make the same mistake :) 180 and 3.14 were sort of mixed in mind. Thank you so much! =D

Oh I was thinking of ∏ = 180. Is it wrong? =/

But isn't (∏)^2/12 = 2700 and 12/12 = 1 Wouldn't it be sort of equivalent to 2700 - sqrt of 3 (1.73) ? That's what I am trying to understand. What's the difference if I work with 30 instead of ∏/6 ? Is it wrong? =/

But is it wrong if I convert ∏/6 = 30 and work with that?

Homework Statement x3 - sin 2x Find f'(∏/6)The Attempt at a Solution f'(x) = 3x2 - 2 cos 2x f(∏/6) = 2700 - 2 [ (√3/2) ] ---> from 2 [ cos(∏/6)] answer: 2700 - √3 My book has the answer as (∏2 - 12)/12

Yes. Got it! =) Now, I'm going to work on nr. 4

I received a warning for excessive use of colors ?? I thought it was actually better for whoever was reading..to separate each problem by color since I was posting more than one. I didn't do it because I was trying to decorate my post. My bad.

:shy: Hi! 3. h(x) = cos [sec(5∏x)] h'(x) = -sin [ sec(5∏x)][ tan (5∏x)] . 5∏ h'(x) = -5∏sin [ sec (5∏x) tan(5∏x)] How about now? What's the wrong with ∏ on nr 2 ? :/

Homework Statement 1. f(x) = 5 sin (8∏x) 2. g(x) = 4∏ [ cos (3∏x) sin (3∏x)] 3. h(x) = cos [sec (5∏x)] 4. Sketch the graph of each function on the indicated interval, making use of relative extrema and points of inflection. f(x) = 2sinx + sin2x ; [0,2∏] The Attempt at a...