Voiceover:Let's say that
we've got the function F of X
and it is equal to the natural log
of X plus five over X minus one.
And what we want to figure out
is what is F prime of X.
And I encourage you to pause this video
and try to figure it out on your own.
So there's two ways that you can approach this.
I would call one way the easy way.
And the other way, the hard way.
And we'll work through both of them.
The easy way is to recognize
your logarithm properties,
to remember that the natural log
of A over B.
Remember natural log is just log base the number E.
So this is just going to be equal to
the natural log of A
minus the natural log of B.
So if we just apply this property
right over here,
and just simplify this expression,
or at least simplify it from a point of view
in terms of having to take this derivative.
We can rewrite F of X.
We can write F of X as being equal to
the natural log of X plus five
minus the natural log of X minus one.
And when we take the derivative now
with respect to X,
F prime of X,
well this is going to be the derivative
of the natural log of X plus five
with respect to X plus five,
so that's going to be one over X plus five
times the derivative of X plus five
with respect to X.
I'm just applying the chain rule here,
and that's just going to be one.
So this, the derivative of that is that.
And the derivative of this,
well, let's see,
we're going to have a minus sign there
and the derivative of the natural log
of X minus one with respect to X minus one
is going to be one over X minus one
and the derivative of X minus one
with the respect to X is just one
you just multiply this by one,
it doesn't really change the value.
And we're done!
We were able to figure out the derivative of F.
Now what's the hard way
you might be thinking?
Or maybe you did do it
when you tried to approach it on your own.
Well, that's not to simplify this expression
using this property
and just to try and power through this
using the chain rule.
So let's try and do that.
In that case, F prime of X
is going to be the derivative
of this whole thing
with respect to X plus five
over X minus one
which is going to be one over X plus five
over X minus one
times the derivative, times the derivative
with respect to X of X plus five
over X minus one.
This is just the chain rule.
The derivative of this whole thing
with respect to this expression,
times the derivative of this expression
with respect to X.
Just the chain rule.
So let's see,
this is going to be equal to,
let's use some colors here,
this, what I'm boxing off in blue,
that's the same thing as X minus one
over X plus five.
I'm just taking the reciprocal of this.
And then it's going to be times,
and I'll do this in magenta.
No that's not magenta.
So that's going to be times
and I'm going to rewrite it
as the derivative with respect to X
of X plus five times X minus one
to the negative one power.
And I like to write it that way
because I always forget the whole
quotient rule thing.
But I remember the product rule.
So this thing
so let me rewrite it,
I think you already appreciate
why this is the hard way.
So let me write this
this is X minus one over X plus five times
so let's apply the product rule,
The derivative of X plus five,
well that's just one times the second term
times X minus one to the negative one,
so that's one over X minus one.
And then plus,
what's the derivative of X minus one
over negative one.
Well, let's see,
that's going to be, that's going to be
negative X minus one to the negative two power.
So I can say negative one, X minus one
to the negative two.
I can just write it like this.
And then times the derivative of X minus one
with respect to X.
Well that's going to be one.
And then times this.
X plus five.
So actually,
so times, times, X plus five.
So all I did here, product rule.
Derivative, derivative of this
is one times that.
And that gave us that over there.
And then I took the derivative of this
which is this right over here.
Negative one over X minus one
and the, over X minus one squared.
Or you can say negative X minus one
to the negative two power
times this first expression over there.
So there's that derivative.
Now let's see if we can simplify things.
So if we, let's see,
if we were to, if we were to
let me just rewrite everything.
So this is equal to X minus one over X plus five
times one over X minus one,
minus X plus five,
over X minus one squared.
Now let's think about what happens
when we distribute this.
So when you distribute this times that,
this numerator cancels with that denominator
and so we're going to get, going to get,
let's see,
one over X plus five.
And then when you distribute it over here
the X plus five is going to cancel
the X plus five.
And one of the X minus ones
is going to cancel
one of these X minus ones.
And you're going to be left
with just one of those X minus ones
as the denominator.
And so you get F prime of X
is equal to this.
And lucky for us,
we got the same answer either way.
But as we see, the easy way
is much easier than the hard way.