so here's the story i accidentally assigned this question to my calculus 1 students for homework
and i don't think this question is fair at all because first keep in mind that
we are in the product rule and also the quotient rule section and the question
is asking us to find the first and also the second derivative of this function
and of course we can just use the quotient rule to get the first derivative no big deal
okay now we are going to work out the second
due to second derivative but how are we going to differentiate this
again not only it's long but also we do not have the chamber yet
seriously will i check the answer in the back of the book they did provide an answer
i don't know and because i assigned it so i feel like i should do it even though i told my students
that they don't have to do it but let's just see how we can possibly do this and we are not going
to be using the chain rule so here we go of course i have the first derivative right here already
so let's go ahead and just put on f double prime and let's put on a big fraction bar
nothing that should be enough let's do the easy part first square the bottom so we have 1 plus
e to the x squared and we have to square this thanks to the quotient rule now put down the
bottom function on the top right i'm going to write it down right here you guys will see what
1 plus e to the x squared and then we are going to multiply by the derivative of the top all
right the root of the 2x is just a 2 no big deal and the diversity of 2x times e to the x we will
have to use the product rule here we go we add the first function and we multiply by the derivative
of the second which is just e to the x and then we add the second function which is e to the x
times the derivative of the first which is a two okay and now we have to differentiate this again
by using the product rule yeah let me keep this as the first function so we will have minus
x squared times the derivative we have which is e to the x and then we add the second function
times the derivative of the first which is negative 2x all right and that's pretty good
so that's just the first part of the top of the quotient rule now we are going to minus
the top function so let's just go and write that down so it's 2x plus 2x e to the x minus x squared
e to the x and then we multiply by the derivative of the bottom we do not have the quotient rule
so this is what we will have to do so have a look here i am going to write this down
twice so we are going to be differentiating 1 plus e to the x times 1 plus e to the x
so this is how you can get around from not using the chain rule okay product with
action all right keep the first function which is going to be 1 plus e to the x and we multiply by
d derivative of the second which is e to the x all right and then we add the second function which is
this which is 1 plus e to the x and we multiply by the derivative first which is just e to the x
so in fact they are the same thing yeah and then we will just get two of this so two e to the x
times one plus e to the x yeah uh it looks like we should distribute the inside right but no because
e to the x to the first power that's i'm not going to say cool because i don't really think so anyway
i'm going to factor out 1 plus e to the x to the first power and not see what we have
left yeah right here we will just have 1 plus e to the x and keep in mind we
will still have to distribute so let me just put that down and then for this part
um have a look this is 2x times e to the x and this is negative 2x times e to the x
so in fact this and that can slow each other i'm not even going to say pretty nicely
all right so let me just write down the rest which is 2 right here and then plus 2 e to the x
and then minus x squared e to the x so hopefully so far so good
okay now i'm going to multiply this out and then we'll put down the result in a parenthesis here
so 1 times oda which is just all that and that's 2 plus 2 e to the x minus x squared e to the x
and then we do e to the x times all that so it's plus 2 e to the x and then
plus okay e to the x times e to the x is e to the 2 x right so we have 2 e to the 2 x
and then this times that so again we will just have minus x square e to the 2x okay
and then for the second part here we'll take this multiply with all that and keep in mind
we also have to distribute so we will have minus first and then this times that is 4
x e to the x and then minus and then we have 4 again x
e to the 2 x because we have 2 e to the x of e to the 2 x and lastly this is going to be a plus
the bottom yeah fourth power so that's pretty good one plus e to the x to the first power ah
this and that can cancel we get the third power all right and we'll just compile like terms on
the top and then i will have to check the answer because i don't have the answer remember there
even though i tried to work out earlier so let's see uh on the top
this is just an only number right so i'm going to write down a 2 right here
the x and then e to the x and see anything else that has just e to the x no so we just have plus
4 e to the x do we have any x e to the x just this right but this is not the same as that so the next
thing that we do will be this so i will just say minus 4 x e to the x so far so good and then next
x squared and e to the x right this is not that because this is e to the 2x
and this right here is which is whole 1 minus x squared e to the x so this is also done
okay let's look for e to the 2x this is the only one that has e to the 2x so we'll
just write down plus 2 e to the 2x so this is done and then x e to the two x which is this
z sec for this we have minus four x e to the two x and lastly oh wow look at that this is
like a small bonus what do i do with this let's put on blue right here so this part
and that part can be combined wow negative x square uh e to the two x and two of them
right so it would be positive so just put on plus this is a 2x plus x squared and then e to the 2x
and then all divided by this guy one plus e to the x and then raised to the third power
and let me just double check before i box the answer or anything
okay yes i got this one right this time all right well i do have to tell you i did
practice this beforehand earlier but i just want to provide a better organization for all the steps
so hopefully this is clear and maybe with this being done i can reassign this question
to my calculus 1 students for homework let me know what you guys think though that's it