Question

d/dt [(1+t)(1+2t)(1+3t)(1+4t)]=? where t=0.

Answer 1

Product rule within product rule.

Answer 2

isn't it too long ? are there any short way ?

Answer 3

\[[a \frac{ d }{ dt }b+b \frac{ d }{ dt }a] + [c \frac{ d }{ dt }k+k \frac{ d }{ dt }c]\]

Answer 4

Its shorter than using the definition of a derivative and solving it using limits. Yes?

Answer 5

@anewbie do u know any properties of fourth order polynomial?

Answer 6

i don't know exactly but my high school teacher did like it. for example, t is 0 so all t's goes. and only constant numbers left. but i don't remember all things. that's what i'm trying to say with short way.

Answer 7

@MathsIsMyMind

Answer 8

either you can use method said by abbot above
or
you can expand the polynomial and apply d/dt and then t to zero but you shouldnt apply t equals zero in the beginning

Answer 9

i mean after derivation. @MathsIsMyMind ok thanks

Answer 10

@abb0t thank you too

Answer 11

Pleasure!

Answer 12

@anewbie do you have answer? I want to know, too. I don't know the method those guys give you.Please, post if you have the answer.

Answer 13

if you are not sure you can get the answer by a "new " method to you, just ask them for more explanation. we are here for being help, no need to feel shameful on our knowledge. ok, I have to go now but I will be back later to know about the answer when you apply the new method or just expand the polynomial as usual. take care

Answer 14

i made product rule each two of them but i'm not sure if my process is true. i don't know.

Answer 15

I think its the easiest way to first multiply all the terms to get 4th order polynomial and then diferentiating it and finally then make t equals zero

Answer 16

simply expand it and solve it?
\[
{d\over dt}(1+3t+2t^2)(1+7t+12t^2)\\
\quad=(1+3t+2t^2)(7+24t)+(3+4t)(1+7t+12t^2)\\
y'(0)=(1)(7)+(3)(1)=\boxed{10}
\]

Answer 17

that's okeyy.. thank youu :)

Answer 18

hmm!!! at the end up?? wait for a long time and no new to study??? !!! anyway, being patient is the hardest lecture.