Question

differentiate

d/dt (a + (2bt/(1-bt^2)))

need help with the rules.. :/

Answer 1

\[\frac{ d }{ dt } [a+\frac{ 2bt }{ (1-bt ^{2} )}]\]

Answer 2

\(a,b = const \implies (a)' = 0\) so you're off to a flying start....

Answer 3

ya.. and i now the (1-bt^2)^2 will be in the denominator, * -2bt, but I don't know what to do with the numerator

Answer 4

for \(\large \frac{ d }{ dt } [\frac{ 2bt }{ (1-bt ^{2} )}]\), most people would use the quotient rule \(\left( \dfrac{u}{v}\right)^{\prime} = \dfrac{u'v - uv'}{v^2}\)

Answer 5

ahhhh okay... thats what i was looking for. i need a book with all these dang cal rules.

Answer 6

or you can re-write as \(\large \frac{ d }{ dt } [ 2bt (1-bt ^{2} )^{-1}]\) and use product rule....

Answer 7

can you give me the product rule by chance so i can write it down.

Answer 8

\((uv)' = u'v + uv'\)

Answer 9

here's a very comprehensive cheat sheet from a very well regarded source
http://tutorial.math.lamar.edu/pdf/Calculus_Cheat_Sheet_All.pdf

Answer 10

Thank you, appreciate it. I know I should remember all this stuff already, which is why it probably isn't in the book.

Answer 11

good luck:-)

Answer 12

do you see my error?? @IrishBoy123

Answer 13

Ahh you just missed a negative sign on the bottom

\[\large 2bt(1 - bt^2)^{-1}\]
\[\large u'v + uv' = 2b(1 - bt^2)^{-1} + 2bt(\color \red{-}(1-bt^2)^{-2})(-2bt))\]

Answer 14

sorry zona and thanks john

i was occupied over on physics....

Answer 15

Yeah no problem, easy to get occupied over there...physics is that great after all :D

Answer 16

ohhh right bring the -1 in front then subtract 1 from the exponent... silly mistake. thank you guys.