Let's take a look at an example of
determining the first
and second derivative
of an implicit function using
implicit differentiation.
Notice how this equation
is not solved for y,
that's why we'll apply
implicit differentiation.
While we could solve this for y,
we will use implicit differentiation.
So to determine the derivative
of both sides with respect to x,
so from this, we can determine
the first derivative.
To differentiate the derivative
of x-squared with respect
to x would be two x plus,
now to differentiate x times y,
we have to apply the product rule,
so we'll have the first
function, which is x,
times the derivative of the
second function which is y,
the derivative of y with
respect to x would be
one times dy/dx, remember
this is in terms of y,
we have this extra factor of dy/dx,
because we're applying the chain rule,
plus the second function, which is y,
times the derivative
of the first function,
the derivative of x would be one.
And on the right side, the
derivative of four would be zero.
So to determine the first derivative,
we need to solve this for dy/dx.
So we'll go ahead and move two x and y
to the other side of the equation,
so we'll have x times dy/dx
equals negative two x minus y,
and now we can divide both sides by x,
so the first derivative, or
dy/dx, is going to be equal to
negative two x minus y divided by x.
Now if we wanted to, we could factor out
a negative in the numerator,
so we have dy/dx equals
negative of the quantity
two x plus y divided by x.
Now this is only half of the problem.
We also want to determine
the second derivative
of y with respect to x,
so now we'll take the
derivative of the derivative
to determine the second derivative.
So now we'll differentiate both sides
of this equation with respect to x.
And since we'll have to
apply the quotient rule here,
I'm going to go ahead and use this form
of the first derivative,
so the numerator is
negative two x minus y divided x.
So the derivative of the
derivative with respect to x,
would be our second derivative,
d-squared y, all over d x-squared equals,
and here's where we'll have
to apply the quotient rule.
So starting with the denominator,
we'll have the denominator squared,
and then the numerator
will have the denominator,
times the derivative of the numerator,
which would be negative
two minus one times dy/dx,
or just dy/dx, remember
because we're taking
zero with respect to x,
and this term has a y,
minus the numerator
times the derivative of the denominator
which would just be one.
Now we have to perform some algebra,
and perform a substitution
for the first derivative.
Let's go ahead and distribute here,
we'll have negative two
x minus x times dy/dx,
but remember dy/dx is this quotient here,
or this quotient in yellow,
let's go ahead and use the form in green.
And then here we'll
have plus two x plus y.
The nice thing about this product here,
is these x's simplify out,
so now let's clear these parenthesis
and we'll combine like terms.
We'll have x-squared in the denominator,
and we'll have negative two x,
here we're going to have plus two x,
then we'll have plus y
plus two x plus y.
So let's simplify this again.
These are opposites, so
it looks like we'll have
two x plus two y in the numerator.
Now this would be our second derivative,
but a lot of times you will see this
expressed in a different form,
so lets continue this on the next slide.
Let's go ahead and factor the
numerator, factor out the two.
So again, this is our second derivative,
but what you'll notice is if we multiply
both the numerator and denominator by x,
you would have the second
derivative equal to
two time the quantity x-squared plus x y,
all over x-cubed, and
the reason why you might
see this done is notice
that x-squared plus x y
is part of the original equation.
The original equation was given as
x-squared plus x y equals four.
So using the given equation,
we can replace x-squared
plus x y with four,
which does simplify our second derivative.
So we have eight all over x-cubed
for our second derivative.
So sometimes it can be a little tricky
if you're trying to match an answer shown
in the back of a textbook or
in an online homework system,
so hopefully you found this helpful.
I think I might be easy
to stop right here,
and forget about trying
to perform this step here
to simplify the second
derivative as much as possible.
