Derivatives of Natural Logs (ln) Example 3
Hi, everyone.
Welcome back to integralcalc.com.
Today we're going to be taking the derivative of the function f of x equals
ln or the natural log of x times the square root of x squared plus one.
And in order to take the derivative of this function,
we're going to need the product rule formula.
We're also going to need the formula
that tells us the derivative of the natural log of x
which is one over x.
But to take the derivative of this function,
first of all the derivative of f of x is always f prime of x.
And whenever we're taking the derivative of something involving the natural log or ln,
we're going to be referrring to this formula here.
And what this tells us is that
whatever is inside the natural log function,
is going to go straight to the denominator.
We’re going to have one in the numerator
and everything inside that natural log function goes to the denominator.
So in our case,
we're going to take everthing inside the natural log function,
which is everything inside these brackets
and we're going to put it in our denominator.
So one divided by the square root of x squared plus one.
And normally,
that would be all we would have to do if
what was inside our natural log was just x, right?
The formula is dealing with the natural log or ln of x only.
What's inside our natural log function
is much more complicated than just ln of x.
So we do this step here
where we take what's inside the natural log
and we put it in the denominator
and we put one in the numerator.
And that's great.
But now,
we have to attend to what's inside the natural log function.
So in order to do that,
like I said we do this step
but then we multiply by the derivative of what's inside right here.
So the derivative of x times the square root of x squared plus one,
we're going to find using the product rule.
So first of all,
let's convert this x times the square root of x squared plus one to
x times x squared plus one to the one-half.
Remember that the square root of x is equal to x to the one-half
so when we see the square root of x squared plus one,
we can change that to the quantity x squared plus one raised to the one-half power.
We haven't changed anything about that function,
we only changed the way that it's written.
So we're going to transform our function into this right here
and now we need to multiply this part here
by the derivative of this x times the quantity x squared plus one to the one-half.
And inorder to do that,
we need the product rule which tells us that
when we have two functions multiplied together,
f of x and g of x,
we're going to use this formula here to find the derivative
so we'll go through that step by step.
So the first thing we need is f prime of x,
in other words the derivative of our first function.
Well, our functions f of x and g of x are x
and this quantity x squared plus one to the one-half.
So for f prime of x, we're looking for the derivative of x.
The derivative of x is just one.
Let's do this.
One, so we go ahead and write that
and then we are going to multiply it by g of x
without doing anything to it.
So that's going to be x squared plus one to the one-half.
Okay?
Then we add f of x as it is, untouched
so we go ahead and write that x in.
Right? From here again.
And then we multiply it by the derivaitve of g of x.
Well, g of x is the quantity x squared plus one to the one-half.
So the derivative of that,
we're actually going to need chain rule again
to find the derivative of that specifically.
Let's look at down here.
The derivative of x squared plus one to the one-half.
So taking the derivative of this like I said requires us to use chain rule.
The outside function,
remember that with chain rule,
basically what chain rule tells us is that we take the derivative of the outside function first,
leaving the inside function completely alone without doing anything to it.
Then we multiply what we got by the derivative of the inside function.
So our outside function is essentially x to the one-half
and the inside function is x squared plus one.
So imagine if instead of x squared plus one,
you just had x to the one-half.
Right?
We're replacing that x squared plus one with just x.
The derivative of that would be one-half x to the negative one-half.
Right?
Which means the derivative of our outside function right here is going to be
one-half times x squared plus one to the negative one-half.
Okay. So we took the derivative of the outside function
leaving that inside function completely alone.
x squared plus one,
we don't even think about right now.
So we ignore that.
That’s the derivative of the outside function so we'll check that off.
But now we have to multiply that by the derivative of the inside function which is x squared plus one.
The derivative of x squared plus one is just two x plus zero.
The derivative of the constant, one is zero.
So two x plus zero is just two x.
So we multiply by two x.
So this right here is the derivative of g of x.
So we can multiply here by one-half times x squared plus one to the negative one-half times two x.
That's a lot of parentheses.
But that is the derivative of our function g of x.
So we have finished writing our derivative function
and now it's just a matter of simplification
which will be a little bit of undertaking
given how complicated this is.
So to simplify this, we'll have one over x times the square root of x square squared plus one.
I'm going to leave that first term exactly as it is for now.
Inside here,
that one is redundant, we don't have to write it.
So we're just going to have x squared plus one to the one-half.
I’m going to go ahead right now...
remember this formula over here
and rewrite that as the square root of x squared plus one, right?
I haven't changed anything. Just rewrote it.
So x squared plus one to the one-half, that part's done.
Now notice here we're looking at...
let's grab...
we're going to look at this part right here
Okay.So we've got a two in our denominator right here and a two in our numerator
so those two can cancel each other and we'll just be left with plus x squared, right?
And then we've got this multiuplied by x squared plus one to the negative one-half, okay?
So...
let's now do a couple of things.
Instead of...
we'll rewrite this here
x times the square root of x squared plus one
times the square root of x squared plus one plus x squared.
This x squared plus one to the negative one-half,
we can move that to the denominator
by changing the sign on the exponent from a negative to a positive.
So instead of x squared plus one to the negative one-half in the numerator,
we can move that to the denominator
and call it x squared plus one to the positive one-half.
But remember that now that we've transformed it in this way,
we've got x squared plus one to the one-half
where we can rewrite that as the square root of x squared plus one.
So let's go ahead and simplify this now to
one over x times the square root of x squared plus one.
We’ve got this multiplied by the square root of x squared plus one
plus x squared over the square root of x squared plus one.
Okay.
So now, I think we can distribute our...
our term out in front here.
So let's get rid of all these stuff.
So we've got f prime of x equals...
I'm going to go ahead and multiply this first term out through this parenthesis here.
So when I do that,
I'll get the square root of x squared plus one divided by x
times the square root of x squared plus one
when I multiply these two terms together
and then I'll get plus...
let's see.
x squared in the numerator divided by...
we've got x in the denominator
and when we multiply this square root of x plus one by this square root of x plus one,
they will combine to just be x squared plus one, right?
When you multiply the square root of something by the square root of the same thing,
the square root signs go away,
so we've just got x squared plus one there.
So now, we can do some cancellation here.
In our first term,
we can cancel out the square root of x squared plus one
and we're just left with one over x.
And then the second term,
we can cancel out an x from the numerator and denominator
so this one's going to go away completely
and this exponent will go away.
And we're just left with x divided by x squared plus one.
So now if we want to combine these fractions,
what we would do to find a common denominator...
is multiply the second term by x over x, right?
Because the denominator of our other fraction is x
so we multiply that by the numerator and denominator of this fraction
and then... Let's just erase that guy.
and then the first one here.
To get a common denominator with the other fraction,
we'll multiply by it's denominator x squared plus one
divided by x squared plus one
and when we do that,
we'll get x squared plus one over x squared plus one
times x
and then we're going to add to that x squared and again x times x squared plus one.
And now as you can see,
we've got the same denominator.
So given that we have the same denominator,
we can combine the fractions,
we have x squared plus one plus x squared in our numerator
so that's going to be two x squared plus one in the numerator
divided by x times the quantity x squared plus one
and that's really as far as we can simplify our fraction.
So that'll be our final answer.
We'll go ahead and write it formally as f prime of x
which is the derivative of f of x.
And it's equal to two x squared plus one divided by x times the quantity x squared plus one.
And that's it.
That’s our final answer.
I hope that video was helpful to you, guys
and I will see you in the next one.
Bye!