the second week of calculus has come to a close. I understand the simeple concepts of the derivitives like the quotient rule vu^1-uv^1/v^2 and the product rule uv^1 + vu^1 and i even understand the sin, cos, tan, etc. rules as well. My biggest problem is that when you have a big problem and you have to do multiple derivitives i get a little lost. For example one problem says ln 3square root x-1/x+1. My first thought would be to treat it as a natural log 1/u X u^1. So i would make it look like 1/(x-1/x+2)^1/3. but i don't understand what u^1 would be or if i need to look at it some other way.
Another one i didn't really understand was -square root of x^2 +4 /2x^2 - 1/4 ln (2+ square root of x^2 +4/x). So i see from looking at it, it is the quotient rule and there is a square root so you have to make it into a fraction.. what confuses me if when i get to the ln part. Simplifying this was also very hard and i catch trouble with that as well. I just need some help breaking down the problem and looking at everything that needs to be done. I know the formulas and how to set them up, i just don't know exactly when to set them up, if that makes any sense.
Other than that, i have a good grasp on the limit stuff and the vertical and horizontal asymptotes for the test wednesday, i just need some help going over the big derivitive problems before we take the test. The only other thing about derivitives that i don't have a big grasp on yet is the e problems. y= e^x-e^-x/2. I see that the quotient rule is here but i don't know what needs to be done with the e's.
I think i will have a handle on the derivitives once someone helps me understand how to break down the problems and know everything that needs to be done in the problem. So all in all, my second week of calculus was not too bad, but i'm just a little confused at times.
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Trina just a little advice if you say u^1 is u prime it will confuse everyone because u^1=u
ReplyDeleteit would be easier to say u'
Trina all you do for the e's is use the formula e^u X u'. An example of how to use it is if you have e^3x. Then you have
ReplyDeletee^3x X(3)which then simplifies to 3e^3x.
hope this helps
okay, for the problem that you mentioned up there, first you need to use the quotient rule. after you take care of that, you use the formula dylan just gave, e^u X u'. after you plug the problem into the formula, you just simplify. I'm pretty sure that's all you have to do. hope this helps.
ReplyDeletethanks you guys i understand it now.. thank you :)
ReplyDelete