Welcome to a lesson
on basic implicit differentiation.
In order to understand implicit differentiation,
we must have a good understanding of the chain rule.
Let's first review the chain rule,
by determining the derivative of the quantity
four x squared plus x raised to the third
with respect to x.
Notice how we have a composite function here
and therefore we must apply the chain rule
to find the derivative.
Looking at our notes below,
here we have the basic power of differentiation
and here we have the extended power
of differentiation where u is the inner function.
Let's notice for our derivative,
the inner function u
is equal to the quantity four x squared plus x.
And therefore u prime, which is the same as du/dx,
equals the derivative of four x squared plus x
with respect to x.
Which is eight x plus one.
So if we wanted to,
we could rewrite this as the derivative
with respect to x of u to the third.
Which is equal to three u squared times u prime,
or three u squared times du/dx,
which in this case is equal to three
times the quantity four x squared plus x squared
times the quantity eight x plus one.
This is the same idea we need to apply
when performing implicit differentiation
and the term uses a variable different than
the input variable.
For example, here we're asked to find the derivative
of y cubed with respect to x,
where the term is a y term
and the input variable is x.
And therefore we need to apply the chain rule
in order to find this derivative
because we are assuming y is a function of x.
So to show some extra work here,
if we wanted to we could let u
be equal to y
and therefore y prime,
which remember is equal to du/dx,
is really the derivative of y with respect to x
which is dy/dx,
often just written as y prime.
Just as in this first example,
we could rewrite this as the derivative of
u cubed with respect to x,
which we already know is equal to three u squared
times u prime.
But here, three u squared times u prime is equal to
three y squared times y prime
or three y squared times dy/dx.
We often use dy/dx instead of y prime
just because y prime is easily confused with y to the first.
We normally won't show this much work though,
when a term is given using a variable different
than the input variable,
we differentiate like we normally do
and then we have an extra factor of the derivative.
Notice how this is really just a derivative
of y cubed with respect to y times dy/dx.
Next we have the derivative of two to the fifth
with respect to t.
Notice in this derivative the term is a t term
and the input variable is also t,
and therefore the chain rule is not required.
We differentiate as we normally do.
So the derivative of two to the fifth with respect to t,
is just ten t to the fourth.
But I do want to point out,
that if we did apply the chain rule,
it would not affect the derivative.
For example, if we did let u be equal to t,
u prime is really just du/dt
and the derivative of t with respect to t is just one.
So multiplying by one of course,
would not change the derivative.
Next we have the derivative of two s to the fifth
with respect to t.
Notice how here though, we have an s term,
and the input variable is t.
And therefore we need to apply the chain rule
where we're assuming s is a function of t.
So this derivative is equal to the derivative of
two s to the fifth with respect to s times ds/dt.
Well the derivative of two s to the fifth would be
ten s to the fourth,
and then we have times ds/dt.
Again we must apply the chain rule
because we're assuming s is a function of t,
so if we did let u equal s,
u prime would be equal to ds/dt.
Now that we understand if we're given a term
that uses a variable different than the input variable
we must apply the chain rule,
let's go over the formal rules
on how to perform implicit differentiation.
For these notes, we're assuming we're given an equation
that uses the variables x and y,
where x is the input variable and y is the output variable.
So step one, we differentiate both sides
of the equation with respect to x.
Step two, we apply the rules of differentiation
as necessary.
Any time an expression involving y is differentiated,
dy/dx will be a factor in the result,
again because of the chain rule.
Step two, we isolate the dy/dx terms on one side
of the equation.
Step four, if there's more than one dy/dx term,
we factor out the dy/dx.
And finally step five, we divide both sides
of the equation to isolate dy/dx.
And let's look at one example before we go.
We're asked to differentiate with respect to x.
So the first step,
is we differentiate both sides of the equation
with respect to x.
Now focusing on the left side.
We know the input variable is x
and because this is a y term, and so is this,
we differentiate these terms.
We'll have an extra factor of dy/dx,
and notice how these two terms are x terms,
and therefore we do not have an extra factor of dy/dx.
So the derivative of y cubed with respect to x
is three y squared times dy/dx plus
the derivative five y with respect to x
is five times dy/dx minus
the derivative of two x to the fourth with respect to x
would be eight x cubed minus the derivative x
with respect to x is one
equals the derivative of 12 respect to x is zero.
The next step is to isolate the dy/dx terms.
So we add eight x cubed and add one to both sides.
Which gives us this equation.
Notice how we have two dy/dx terms on the left side,
so now we factor out dy/dx and those two terms.
Giving us dy/dx times the quantity
three y squared plus five on the left.
And our last step is to divide in order to solve for dy/dx.
Which gives us our derivative.
I hope you found this helpful.