We'd like to take the derivative of a function
raised to another function. In this case,
it is a very simple X raised to the X but
it could be something more complicated like
sine raised to the cosine. How can we take
the derivative like this? We cannot use the
X to a number rule the power rule because
we do not have a number up here at the top.
We cannot use the E to the X rule because
even though we have A on the top we do not
have a number on the bottom. Neither one of
these is going to work. We need to come up
with something completely new and what we
are going to do is we are going to end up
doing logarithmic differentiation. We are
going to take the log of both sides and now
we can use that log rule that we talked about
before, log of A to the B equals B log A to
bring down an X that is in the exponent to
and from the log. Now, this looks a little
bit like the power rule, make sure you do
not get confused. This is log rule from precalculus
as opposed to the power rule which gives you
N X to the N minus one and that is a derivative.
This here is not taking a derivative it is
just doing algebraic manipulation. We have
natural log of Y equals X times natural log
of X. Now we know how to take the derivative
of natural log of Y and we know how to take
the derivative of X and we know how to take
the derivative of natural log of X. We can
do D/DX of both sides. The derivative of natural
log of Y is one over Y times DY/DX and the
derivative of X times natural log of X we
need to use the product rule. The derivative
of X is 1 times natural log of X unchanged
plus X unchanged times the derivative of natural
log of X which is one over X. Now we can solve
for DY/DX. We have natural log of X plus notice
the X times one over X is just 1. The natural
log of X plus 1 times Y and then our last
step when doing logarithmic differentiation
is to replace that Y with the original function,
replacing the Y with X to the X. We have now
taken the derivative of X to the X. Anytime
you have a function raised to a function you
are going to have to use logarithmic differentiation.
When you are doing logarithmic differentiation
to make a really complicated functions simpler
that is a choice. You really could use the
product rule and the quotient rule and power
rule as many times you need to but whenever
you have a function raised to a function like
sine of X to the cosine of X this is only
way to do this problem. Again what would we
do we would write Y equals sine of X to the
cosine of X and we want to take the natural
log of both sides. Now there is another way
of actually notating this. I am going to show
you this in this problem instead of actually
taking the natural log of both sides and using
implicit. I am going to take E raised to the
LN of sine of X to the cosine of X and the
advantage of this is that it does not use
implicit differentiation we can actually do
it in the middle of a more complicated problem
like the one I am going to show you in just
a minute. What does this do for us? Well now
we've got Y equals E and we have L of something
raised to something. We can use that power
rule and bring it down, we have cosine of
X times natural log of sine of X. Now what?
Well now, we have a something complicated
looking but something we can do with just
the chain rule and the product rule and the
chain rule again. Taking the derivative here
the outside must thing E, E to the something
is something times the derivative of the something.
Now we are going to use the product rule because
we have two things being multiple here cosine
and natural log of sine of X. We have derivative
of cosine which is negative sine times the
natural log of sine unchanged plus we are
running out of room. I am going below, cosine
unchanged times the derivative of natural
log of sine of X. Now remember derivative
of natural log is one over. We get one over
sine of X times the derivative of sine of
X because the chain rule cosine of X. There
we have taken the derivative of sine raised
to the cosine of X we have used the trick
of log to bring down the exponent but we haven't
had to do implicit differentiation and we
were still able to find the answer. Let's
try one more example. This is X raised to
the natural log of X plus 8 X squared and
this is an example like what I was referring
to in the last problem where you have something
where you need the log and a function raised
to a function but it is inside the context
of a bigger problem. Here how would you do
this? You would write F of X equals E raised
to the natural log of now here is our function
raised to a function X to the LN of X plus
8 X squared. Now the fact that we have the
8 X squared there does not really hamper us
at all as opposed to if you try to do the
natural log of both side here if you had Y
equals X to the LN of X plus 8 X squared and
you take natural log of both sides of that.
Now we can't actually apply that log rule.
We can't bring this out front because it is
not the whole thing inside the log is not
raised to that power. As soon as you have
a plus sign inside a log like that you can't
use any more rules, log rules to simplify,
this does not work at all. Down here back
to our E to the LN with dealing with this
now that makes sense to do right because E
and LN undo each other really haven't done
anything to this function X to the LN of X
is the same as E to the LN of X to the LN
of X. Now we can just use regular derivative
rules to take the derivative of this equation
as soon as we rewrite it one more time. The
thing we need to do still is to take this
LN and bring it out front. We get E this LN
out front times and the LN that was left from
before which was just the LN of X that is
kind of confusing we have LN of X times LN
of X plus our 8 X squared and now we are just
going straight to F prime using our regular
derivative rule product, quotient, chain,
no implicit is needed. The derivative of E
to the something is E to that something. This
is E to the LN of X of the whole thing squared
because there is two of them right multiplied
together and note again you cannot use the
natural log rule of bringing down the power
on this one because the 2 is outside of the
log it is not inside the log. We have E to
the natural log of X squared times the derivative
of natural log of X squared, so that's times
2 natural log of X because the outside function
here is something squared. It is 2 to the
first power but now I have to take the derivative
of that inside. The derivative of natural
log of X is one over X. We've done two chain
rules here actually the derivative E to the
something is E to that something times the
inside there natural log of to the second
power becomes two natural log to the first
power but now I have to take the derivative
of natural log and we get one over X and now
we are going to add to that derivative of
8 X squared which is just 16 X and there we
go. There is a using the power of log to take
a derivative of a function raised to a function
added to some other function.
