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Calculus review! |
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| Dec17-03, 03:31 PM | #1 |
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Calculus review!
Hi guys, I have a calculus test tomorrow and it'll be great if I can recieve some answers + explanations! my review sheet is attached. if this is breaking the rules im sorry. [g)]
hope this actually helps some other people too [:D] |
| Dec17-03, 03:33 PM | #2 |
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for #1
i got 3x^2-sin(xy^2)*2xy [dy/dx]*y^2 whadaya thinks? [:D] |
| Dec17-03, 03:36 PM | #3 |
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for #2) its log(x)/log4 b)then graph it with my TI89 then find the deriv @ .4
at work right now so no calculator on me =) btw: does the TI89 have a test function == like the ti86? |
| Dec17-03, 03:40 PM | #4 |
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Calculus review!
#3) f'(x)=2+cosx so we have to prove f'(x) is constantly positive or negative.. but how would we approach that?
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| Dec17-03, 03:46 PM | #5 |
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Recognitions:
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| Dec17-03, 03:53 PM | #6 |
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| Dec17-03, 04:24 PM | #7 |
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Recognitions:
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So, if the minimum value of cos(x) is -1, what is the minimum value of 2+cos(x)?
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| Dec17-03, 04:35 PM | #8 |
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lol wow i feel so dumb [*(]
[6)] thanks onward to the next question after i finish some quick job for my boss. [:D] |
| Dec17-03, 04:42 PM | #9 |
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i got x-4=y^3+y
so turn my ti89 to parametric mode and y1 = t^3+t y2= -1 right? (since its g(3) 3-4 = -1) then i trace for a zero? |
| Dec18-03, 11:10 AM | #10 |
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just came back from my test and i got confused on a question... find the deriv of x+sin(xy) i got (-sec(x)-y)/x but the calc gave me 1 [o)]
thanks to all the people that replied and (attempted to) help [:D] |
| Dec18-03, 11:10 PM | #11 |
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d/dx{x+sin(xy)} = 1 +cos(xy)[xdy/dx+y]
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| Dec19-03, 10:56 AM | #12 |
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oh my goddd what did i do!?! i messed up.. oh wells [:)]
thanks himanshu121 for reminding me of my errror |
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