Welcome to a video
on logarithmic differentiation.
And the goal of this video
is to use this technique called log differentiation
to help differentiate nonlog functions.
This technique can be useful for some functions
that might otherwise be very challenging to differentiate.
And here are the steps
to perform logarithmic differentiation.
Step one, we'll take the log of both sides of the equation.
Step two, we'll apply
the appropriate properties of logarithms.
Step three, we'll differentiate the equation.
Step four, we'll solve for dy/dx.
And then step five,
we'll use the original function to substitute for y.
But before we get started, let's go ahead and review
some of the properties of logarithms.
These first three are going to be key:
the product property of logs,
the quotient property of logs,
and the power property of logs.
And notice we will be using natural logs.
And then one more thing to review,
the derivative of natural log u
is equal to 1 over u times u prime,
which is equal to u prime divided by u.
Okay, so let's take a look at a couple examples.
We want to find dy/dx.
Now, this is an explicit function,
but in order to find the derivative using the quotient rule,
we'd also have to apply the chain rule and it would be
very challenging to keep all of that straight.
So we'll apply logarithmic differentiation on this function
to find its derivative.
So we'll take the natural log of both sides of the equation.
I'm also going to rewrite this as x squared plus 1.
Notice how, when I did this,
I also rewrote the square root in rational exponent form.
Now, we can apply the property of logs,
meaning because this is a quotient,
we can write it as a difference of two logs.
Let's go ahead and do that.
And now we can apply the power property.
So we would have 2 natural log
x plus 2 minus 1/2 natural log
x squared plus 1.
Now we're going to take the derivative of both sides
of the equation with respects to x.
So on the left side, the derivative of natural log y
would be 1 over y times dy/dx.
Remember, this is in terms of y,
so we have a factor of dy/dx equals
2 times 1 over u times u prime.
Well, u prime would be 1.
So we'd have 2 times 1 over x plus 2
or 2 over x plus 2
minus 1/2 times u prime over u.
Well, u prime would actually be 2x,
and, of course, u would be x squared plus 1.
So again, here's the u,
so there's u prime and then there's a u.
We can also see this simplifies here.
Now all we have to do is solve this for dy/dx.
So we'd have dy/dx equals y
times the difference of these two fractions.
Now, depending on your instructor,
you may be able to leave your answer in this form.
However, we know what y is equal to,
because that was the given function.
So we could replace y with this quotient
to have everything in terms of x.
Let's go ahead and show this on the next screen.
Let's go ahead and replace y with a function.
So we'd have dy/dx equals the quantity
x plus 2 squared over the square root of x squared
plus 1 or the quantity x squared plus 1
to the 1/2 power
times the difference of these two fractions.
Let's go ahead and subtract these.
Let's first get a common denominator.
We can see our common denominator would be
x plus 2 times x squared plus 1.
Let's do some work on the side here
to determine our numerator.
We'd have to distribute and then combine like terms.
So we would have 2x squared plus 2.
Here we'll distribute a negative x.
We'll have minus x squared minus 2x.
So when we combine these terms,
we're going to have 2x squared minus x squared.
That's 1x squared minus 2x plus 2.
Now we can simplify further.
Notice there's a common factor here of x plus 2.
So this simplifies and this changes to x plus 2
to the first power.
So dy/dx is equal to x plus 2
times the quantity x squared minus 2x plus 2
over, now if you look at the denominator,
we have a common base of x squared plus 1.
This is to the 1/2 power,
and this is to the first power.
Remember when the bases are the same
when you're multiplying, you add your exponents.
So 1/2 plus 1, that would give us
the quantity x squared plus 1 raised to the 3/2 power.
Again, 1/2 plus 1 gives us 3/2.
Okay, let's go ahead and take a look at one more example.
Now we want to differentiate y equals x
to the power of 2 divided by x.
So we'll take the natural log of both sides.
Now we can apply the power property.
So we would have natural log y
equals 2 divided by x
times natural log x.
We may want to write 2 divided by x
as 2x to the negative 1 power
since we know we'll have to find the derivative
of both sides of the equations.
Let's go ahead and do that for the natural log y
equals 2x to the negative 1 power times natural log x.
Now we'll find the derivative of both sides
with respects to x, and this will require the product rule.
So on the left side,
we'll have 1 over y times dy/dx
equals the derivative of the right.
So we have the first function
times derivative of the second.
That'd be 1 over x plus the second
times the derivative of the first.
That would be negative 2x to the negative 2.
Let's see what we can do with this.
Remember, we could make this a positive exponent
by moving it down and the same thing here.
So on the right side,
we do have a common denominator of x squared.
So we have 1 over y times dy/dx equals,
this would be 2 over x squared
minus 2 natural log x all over x squared.
Now, the next step would be to multiply both sides
of this equation by y.
Let's go ahead and do this
and move it over to the next slide.
Again, we'd multiply both sides of the equation by y.
So we have dy/dx equals y times,
I'll go ahead and write this as a single fraction.
We have 2 minus 2 natural log x over x squared.
Now, remember we can replace y with the original function.
So dy/dx would equal x to the power of 2
divided by x
times 2 minus 2 natural log x
divided by x squared.
Next, let's go ahead and factor out the common factor
of 2 over x squared.
And then lastly,
and then we're getting a little fancy here with the algebra,
but we have 2.
Now, two bases are the same.
And since we're dividing here,
we could subtract our exponents.
So this could be written as x to the power of 2
divided by x minus 2,
all times 1 minus natural log x.
So there's a lot of fancy algebra going on here.
Really depends on how important it is
for you to get your answer to match
your online homework or the back of the book
or the preference of your instructor.
Personally, I'm okay with leaving it
in this form or this form,
but some instructors may emphasize using your algebra skills
to manipulate the form.
I hope you found these explanations helpful.
Thank you and have a good day.