The topic of today's lecture is 
Implicit and Logarithmic Differentiation.
Up to this point, we've seen how 
to differentiate explicit functions.
These are functions whose defined equations are solved for y.
We will now learn how to differentiate Implicit functions.
These are functions whose defined equation is not solved for y.
Throughout the rest of this lecture,
we will use the Chain Rule time and time again
so if you need to take some time and review the Chain Rule.
Let's now see how we implicitly differentiate.
To differentiate these relations implicitly,
we assume that y is a function of x.
Then, we differentiate both sides of the equation with respect to x,
making sure that we use 
the Chain Rule on occurrences of y.
We'll also need to use all the previously learned differentiation rules
when differentiating both sides of the equation.
Finally, after you differentiate 
both sides of the equation,
bring y prime to one side of the
equation and solve for y prime.
Here's the first example.
Differentiate the following relation
implicitly with respect to x.
Four x cubed minus two x y squared plus y to the fifth plus six is equal to zero.
Notice that the term
negative two x y squared, 
and y to the fifth
will require  us to use the Chain Rule as they involve occurrences of y.
 
Here's the solution:
Differentiating the left hand side with respect to x,
using the Product Rule and the Chain Rule,
we see that the derivative 
of the left hand side is
four times three x squared minus two y squared
minus two x times two y times y prime,
where we use the Chain Rule,
plus five times y to the fourth times y prime,
again where we use the Chain Rule,
plus zero
is equal to the derivative of the right hand side with respect to x,
which is zero.
Now, we'll solve for y prime.
Bring the terms of y prime to one side of the equation
and throw the terms without y primes on the other.
Doing this, we see
negative four x y times y prime plus five y to the fourth times y prime
is equal to negative twelve  x squared plus two y squared.
Factoring out the y prime, we see
y prime times the quantity negative four x y plus five y to the fourth
is equal to negative twelve x squared plus two y squared.
Solving for y prime, we see that y prime is equal to
negative twelve x squared plus two y squared divided by
negative four x y plus five y to the fourth.
Let's look at the next example.
Find the slope of the curve
y is equal to two times the quantity of x plus y to the one third
at the point (four, four).
Here's the solution:
We need to find y prime so we implicitly differentiate the equation.
Taking the x derivative of both sides of the equation,
we see that y prime is equal to
two times one third times the quantity x plus y to the power negative two thirds
times the quantity one plus y prime.
Writing this as a fraction, we see that y prime is equal to
two times the quantity one plus y prime
all divided by three times the quantity x plus y to the power of two thirds.
Manipulating this to solve for y prime, we see that y prime times
three times the quantity x plus y to the power two thirds minus two
all divided by three times the quantity x plus y to the two thirds
is equal to two divided by three times x plus y to the power two thirds.
Therefore, solving for y prime,
we see that y prime is equal to
two divided by the quantity 
three times the quantity
x plus y to the power two thirds minus 2.
Finally, to find the slope at the point (four,four),
we plug in x equals four, and y equals four.
Doing so, we see that the slope of the given curve is y prime
is two divided by (three times 
eight to the two thirds minus 2)
which simplifies to one fifth,
so the slope of this curve at the point (four, four) is one fifth.
So in order to implicitly differentiate an equation,
we apply the ordinary rules of the
derivatives on both sides of the equation
and use the Chain Rule at  every occurrence of y.
Using Implicit Differentiation,
we can now find the slope of the tangent line of curves
whose defining equation cannot be uniquely solved for y.
Furthermore, as inverse functions are defined by
interchanging the independent and dependent variables,
Implicit Differentiation can be used to find the derivatives of inverse functions.
Before we see some more examples on Implicit Differentiation,
let's make a few comments on them.
Here are a few remarks about Implicit Differentiation.
First, when implicitly differentiating a relationship,
the derivative y prime will generally depend on both x and y.
Second, Implicit differentiation can be used to find
the derivative of inverse trigonometric functions.
and more generally, the derivative of inverse functions.
Finally, Implicit differentiation can be
applied as many times as needed
to find the higher order derivatives.
Once you have a formula for y prime, you can find the formula for y double prime
and then plug in the formula you've already found for y prime
to find higher order derivatives.
Here's the next example:
Find d squared y  / d x squared if x cubed plus y cubed is equal to one.
To do this, we have to implicitly
differentiate the equation twice.
Here's the solution:
If we take two x derivatives of both sides of the equation, we see that
d squared of (x cubed plus y cubed) over d x squared
is equal to d squared of one over d x squared.
Notice that the right hand side of the equation will simplify to zero.
Taking the derivative one time, we see that
d of (three x squared plus three y squared times dy/dx) with respect to x
is equal to the derivative of zero.
Taking the derivative again, we see
six x plus six y times dy/dx plus
three y squared times d squared y d x squared is equal to zero.
Plugging in what dy/dx is, we can now
solve for d squared y d x squared.
Here, we learn how to find the
derivative of a Logarithmic Function.
so we'll find the derivative of y is equal to the natural log of x.
First, notice that y is equal to the natural log of x
if and only if x is equal to e to the y.
This is the inverse function relationship.
Now the equation x equals e to the y
can be implicitly differentiated with respect to x.
So taking the derivative with respect to x of both sides,
we see that d dx of x is equal to e to the y times dy/dx by The Chain Rule.
Solving for dy/dx, we see that dy/dx is equal to one over e to the y.
However, since e to the y is x, we see that dy/dx is one over x.
That is, the derivative of the natural log of x is dy/dx is equal to one over x.
So the derivative of the natural log of x is one over x.
Now that we know how to differentiate the natural log function,
let's consider the following example:
Differentiate the following equation with respect to t.
f of t is equal to t times the natural log of t minus two t.
Here's the solution:
We'll use the Product Rule to differentiate f of t.
f prime of t is equal to
one times the natural log of t 
plus t times one over t minus two.
Here we use the fact that the derivative of the natural log of t is one over t.
Simplifying this expression, we see that
f prime of t is equal to the natural log of t plus one minus two
which further simplifies to the natural log of t minus one.
Knowing the derivative of the natural log function,
can help us to find the derivative of the log with an arbitrary base.
Here's the example to see this:
Differentiate the following equation with respect to x:
y is equal to the log base five of (one hundred x plus three).
Here's the solution.
The change of base formula tells us that the log base b of u is equal to
the natural log of u over 
the natural log of b.
This implies that the log base five of (one hundred x plus three)
is the natural log of (one hundred x plus three) all divided by the natural log of five.
So differentiating y, we see that y prime is equal to
the derivative of the natural log of (one hundred x plus three)
all divided by the natural log of five.
Since the natural log of five is a constant, we can pull it out of the derivative
so y prime is equal to one over the natural log of five
times the derivative of the natural log of (one hundred x plus three).
The Chain Rule tells us that this is equal to
one divided by the natural log of five times
(one over the quantity one hundred x plus three) times one hundred.
Simplifying, we see that y prime is equal to
one hundred divided by (the natural log of five times the quantity one hundred x plus three.)
Here are a few remarks on logarithmic differentiation.
In order to differentiate a function,
where the variable x appears in both the base and the exponent,
we first take the natural log of both sides.
Doing this takes the terms with x out of the exponent.
In some cases, this method for differentiation,
known as logarithmic differentiation,
may not be the only option but it could very well be the easiest.
To summarize, to implicitly differentiate an equation,
take the derivative with respect to x of both sides of the equation
use the Chain Rule whenever you have 
a term involving y and solve for dy/dx.
Logarithmic differentiation can be extremely useful
when you have the product or quotient of a function with many terms.
Implicit differentiation is also useful
when trying to find the derivative of inverse functions.
Take some time to familiarize yourself with the techniques above
and try some of the problems we provided for you below.
Good luck practising with your problems, and thanks for listening.
