we will use the definition of derivative to find f'(a)
namely the derivative of f
and some number a and at the moment we can only use the definition of f prime of a
and we have two of them i will show you the first one so i will write down okay f prime
of a by definition this is what i call the definition one this is defined as the limit
but x is approaching to the number that we're looking at here which is a and then
we have f of x minus f of a over x minus a that's what i call the definition one of the derivative
and now we just need to do the work we have to make sure we set our limit equations correctly
let me write it down again x approaching to a for the limit here we are looking at and f
of x which is this so i just need to plug that in for that 3x squared minus 4x plus 1 that's f of x
and then i'm going to subtract f of a and to get f of a i have to plug in a into all the x in the
function but since i'm subtracting the whole c let me open the big parentheses right here first
and i'm plugging a into only x so i will get 3 a squared minus 4
a and we have plus 1 like that and for the denominator we still have x minus a
next this is what we are going to do this is going to be still the limit when x is
approaching to a and let me just write down the denominator x minus a the key right here is that
i'm going to group these two together and let me just show you how are we going to
do it and i'll tell you why in a minute so i'll write down 3 x squared minus 3 a squared
okay i'll group them together we have the negative x minus negative four a in another word
we have negative four x minus minus becomes plus plus four a just like that
we have one minus plus one in other words they pretty much cancel each other
so if you take out the parentheses and then by doing just rearranging it and this is what we have
at the moment and if you plug in a into all the x we are going to end up with zero over zero and we
should somehow see something that we can cancel on the top and with the bottom and to do that
now i will tell you why i want to uh rearrange them because i can factor this by grouping
because we have four terms for this part both of them have the 3 so i can factor out a 3
and i will get x squared minus a squared like that both of this have a 4 but then you have to
maintain the same sign so i'm going to factor out negative negative 4 instead a factor of negative
four and then now open the parenthesis this right here will be x but originally i have a positive
right here a factor or negative so it becomes minus a right here and then notice that this part
x squared minus a squared this is the difference of two squares so once again we can factor this
out and i'm going to put down the three right here and then we'll break this down as x minus a
times x plus a and now we are ready i will write this down as the limit when x is approaching to a
same thing and then the denominator also stays as x minus a but on the numerator this right here is
the first term i'm looking at this as 3 times x minus a times x plus a that's the first term
and notice that for the second term we also have x minus a so we can factor out the x minus eight
i'm going to factor out this i'm this expensive okay and now if you look at the first part we
will have three times x plus a left i factor this out let me open bigger parentheses for
the first part i just have this thing left which is 3 times x plus a 3 times x plus 8.
minus four and then