BAM! Mr. Tarrou. In this video we're going to be introduced to the idea of a second derivative
finding it implicitly okay now how we're going to do that well we're just going to take the
derivative of both sides of the equation with respect to x like we've been doing for this entire
section so we're going to take the derivative of the left hand side and the right hand side
the derivative of x squared with respect to x is going to be bring the power down reduce
the power by one and i'm going to go ahead and show the chain rule for every single
term here even though in this case like with the derivative of the inside function
in what is inside the power of 2 which is x take the derivative of the inside function with respect
to x well that dx over dx is going to cancel that's why there is no x prime on the second
term we are going to bring the power of 2 down so 2 y then to the first paying attention to what is
inside that power of 2 and finishing the chain rule we are going to take the derivative of y
with respect to x so if you hadn't caught on yet that's why we have a y prime and not
y and y we don't have any x primes when we're dealing with equations that involve variables
of x and y and we're taking the derivative with respect to x on the right hand side we have the
derivative of a constant one more time is equal to zero so we have two x plus two y
y prime is equal to zero subtracting both sides by 2x we have 2y y prime is equal to negative 2x and
ultimately we have a y prime is equal to negative x over y great that's step one
and nothing maybe necessarily very interesting there step two we are asked to find
a second derivative implicitly so we're going to oh you know what i'm going to shift back to
the derivative of y with respect to x and we're going to use that notation d y over d
x instead of y prime for our second derivative so on the left hand side one more time we're
going to take the derivative with respect to x of on the left hand side are d y over dx
and we're going to be taking the derivative with respect to x
of our left hand uh right hand side which is again negative x over y
when you look at the notation of a second derivative
the little exponent of 2 if you're not using like a y double prime notation it shows up in different
places on that notation well think about what we have here the derivative of y with respect to
x and now we're going to come along and apply another derivative with respect to x
well it appears that we have like a d times d so our second derivative notation is going to be
kind of like d squared y and in the denominator we have a dx times dx leibniz the mathematician
that created this notation made this notation that we're using in our in a lot of calculus
to make it easier to use the proper notation by if it looks like multiplication if it looks like
division that's how the notation acts even though it's really about the notation of the
of of the calculus and not really maybe truly like multiplication and division
so we have dx over dx times dx is that dx squared so there is how you write if you will y double
prime in that expanded notation d exponent of 2 y over d x squared and on the right hand side
we have of course the derivative of a quotient so that's going to be low
so y d high the derivative of negative x with respect to x is going to be negative 1
minus high or the numerator so minus negative x d low the derivative of
the denominator well so let's just write this as
a y prime now we'll use it as we'll write d y over dx all over the denominator squared now
you'll never be allowed to write a second derivative equation with a first derivative
notation in your final answer so we need to get out the d y dx we need to remove
that first derivative notation and replace it with what it's equal to
so that's one reason why we're doing this video lots of students will
incorrectly leave first derivative notation in a second derivative answer and that is
generally not going to be desirable so we have y double prime i want to go back to that notation
we have a negative y negative times negative is positive so negative y plus now plus x
we're just going to do substitution that's all we're doing we're removing that notation
of d y over d x or y prime and replacing it with what is equal to which is negative x
over y and then that is all over our denominator of y squared
now there's a few ways you can do this we can write negative y over 1
and find a common denominator we can or we can do something to eliminate the division of y
in the numerator which is the inverse math operation to division is multiplication so
what i'm going to do is i'm going to show that i'm multiplying the numerator by y
and the denominator by y and just do a little bit less work eliminate that division of y
because i'm running out of board space and we're going to distribute the y through both
terms in the numerator and then distribute that or multiply that y to the denominator
and we have y double prime is equal to well negative y squared is we have negative y times y
in our second term the division of y and the multiplication of y are going to cancel so
our second term is just going to be minus x squared over y cubed and that's great
we have a second derivative that doesn't have first derivative notation in it now
another reason why i chose this example it's not a really
terribly complicated example to introduce second derivatives it's never a bad idea to introduce
with a fairly simple example before it gets all nasty another thing is this may very well be not
one of the choices say on a multiple choice test it might be quite a bit simpler looking
you see if i take a negative 1 if i factor a negative 1 out of both terms in the numerator
we're going to have and also doing your work horizontally is generally not a good idea but i'm
running out of space i'm going to take a negative out and have y squared plus x squared over y
cubed now if i do that do you see any notate any any potential substitution like why
if i have this sort of complex ish fraction for my second derivative maybe y are all my
multiple choice answers so simple looking and maybe even have a number in it
x squared plus y squared is equal to 16. so it turns out here that when we go to write
our second derivative not only and we have to do this not only do we have to take out
with substitution remove that first derivative notation but you might just find that the answer
lends back and goes back to that original problem and also gives another opportunity
for substitution we're going to take out the x squared plus y squared and
replace it with replace it with 16 because that's what it was defined to be equal to
in the original problem so our second derivative y double prime is equal to negative 16 over y cubed
and i hope that's enough to help you answer every second derivative uh question in your homework
that is going to be finding that second derivative implicitly i'm mr tarrou go do your homework
you