Ask your own question, for FREE!
Mathematics 20 Online
OpenStudy (anonymous):

If f(X)= e^2In x, then f'(3)=

OpenStudy (ribhu):

=e^(lnx^2)

OpenStudy (ribhu):

f'x = 2x

OpenStudy (anonymous):

I don't understand??

OpenStudy (ribhu):

prop of log

OpenStudy (anonymous):

so how do you get f'(3)?

OpenStudy (ribhu):

put x=3 in f'x =2x

OpenStudy (anonymous):

so would the answer be 6:)

OpenStudy (anonymous):

???

OpenStudy (ribhu):

yups i have shown each and every step\[a^{\log_{a}x } = x\]

OpenStudy (anonymous):

Thanks:)

OpenStudy (ribhu):

this is property of log whic's being applied

OpenStudy (anonymous):

The position of a particle moving along a straight line at any time t is given by s(t)=t^3+9t^2-27. What is the velocity of the particle at the time when the acceleration is zero?

OpenStudy (ribhu):

acceleration = s"(t)=0, get a value of t then put in s'(t) which would be velocity

OpenStudy (anonymous):

so would s"(t)=6t+18 ?

OpenStudy (anonymous):

......then set is equal to 0 to get -3??

OpenStudy (ribhu):

yaa there seems something wrong in the question

OpenStudy (anonymous):

why?? -3 is an answer choice?

OpenStudy (ribhu):

time can not be negative

OpenStudy (anonymous):

then how would we get the correct answer?? b/c I'm pretty sure -3 is right......idk though

OpenStudy (ribhu):

-3 comes from s"t = 0

OpenStudy (anonymous):

yes...

OpenStudy (ribhu):

and u've to calculates't at that t at which s"t -0

OpenStudy (anonymous):

I set s''t=0......the answer was -3

OpenStudy (ribhu):

yaa so t=negative not possible

OpenStudy (anonymous):

okay....so do you know what we do from here?

OpenStudy (ribhu):

the quest is probably having a mistake.

OpenStudy (anonymous):

I read over the ? & everything seems right?...do you want to move to the nxt question? I'll just bubble -3 as my answer...:)

OpenStudy (ribhu):

its asking velocity which should be zero at t=-3, velo = -27 from this data

OpenStudy (anonymous):

yes....-27

OpenStudy (anonymous):

?

OpenStudy (anonymous):

r u still there??.....:)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!