We want to determine y prime
and the derivative of the given function.
Notice, this is a pretty complicated function,
so what we can do is take advantage
of the properties of logarithms
to help us determine the derivatives.
So what we can do is take the natural log
of both sides of this equation,
and then we can apply the properties
of logarithms given here, to expand this logarithm.
Let's first eliminate this quotient
by writing this as a difference of two logs.
We would have natural log y equals
natural log of x to the 4th
times the square root of x minus three
minus natural log x plus one to the third.
Now, let's expand this logarithm here.
Since we have a product of two functions,
we can write this as a sum of two natural logs.
So, we'd have natural log y
equals natural log of x to the 4th
plus natural log of the square root of x minus three,
which I'm going to write as the quantity
x minus three to the 1/2,
minus natural log quantity x plus one to the 3rd.
And the last thing we'll do here is
use the power property of logarithms.
So I'll move the exponents to the front
and multiply it by the natural log.
So we'll take this four and move it to the front.
We'll take this 1/2, move it to the front.
And we'll take this three and move it to the front.
So we'll have natural log y equals
four natural log x, plus 1/2 natural log x minus three
minus three natural log x plus one.
So now, I'll take the derivative of both sides
of the equation with respects to x.
So when we differentiate natural log y
with respects to x, we're going to have
one over y times dy dx.
We have to apply the Chain Rule.
So we have one over y times dy dx,
is going to equal four times the derivative
of natural log x.
That will be four over x,
plus 1/2 times the derivative
of the natural log x minus three.
We may be thinking the Chain Rule here,
but if u is equal to x minus three,
u prime would be equal to one.
So the derivative of natural log x minus three
is just one over x minus three.
So we'd have one over two times x minus three here,
minus three times the derivative of natural log x plus one,
which is one over x plus one.
Again if we applied the Chain Rule,
u prime would just be one.
Now, the next step is going to be to add
these fractions on the right side.
So we have to obtain a common denominator,
which will be two, x, x minus three, and x plus one.
So we need to multiply this fraction by two,
x minus three, and x plus one,
on both the top and bottom.
We'll have to multiply the top and bottom
by x times x plus one for this fraction.
And here, we'll have to multiply by 2x
and x minus three on both the top and bottom.
So notice how all the denominators are now
the same, so we can try to combine the numerators.
So we'll have one over y times dy dx.
It's going to be equal to,
a fraction where the denominator is going to be
2x times x minus three times x plus one.
Now, we'll multiply these two binomials
and then we'll multiply that by eight.
So x times x is x squared, times eight.
That's 8x squared.
Then, we'll have plus x minus two x,
that's minus two x times eight.
It's minus 16x.
Then, we have negative three times one.
It's negative three times eight,
so we'll have minus 24.
Here, we're going to have x squared plus x.
Here, we need to be careful because we're subtracting,
so we'll have three times 2x, that's 6x.
6x times x would be minus 6x squared.
And then, we'll have 6x times negative three.
That's negative 18x, but we're subtracting it.
So it's plus 18x.
Now, let's take this to the next slide
and simplify the numerator.
Here, we have an x squared term,
an x squared term, and an x squared term.
So we'll have 9x squared minus 6x squared.
That's 3x squared.
Now, let's look at our x terms.
We have negative 16x plus x plus 18x.
That's going to be plus 3x.
And then, we have minus 24.
Notice how there is a common factor of three here.
So we're going to factor this three out.
But we also have to solve this equation for dy dx.
We'll do that by multiplying both
sides of the equation by y.
Multiply this equation by y.
The ys will simplify out.
So now, multiply the right side by y.
So if this simplifies out, we're left
with dy dx equals y,
which is the original function given here.
So, we'll have x to the 4th times
the quantity x minus three to the 1/2,
all over x plus one to the 3rd,
times, again, we're going to factor this numerator,
so we'll have three times the quantity x squared
plus x minus eight.
All over 2x times x minus three times x plus one.
So again, here we multiply both sides
of the equation by y to solve for dy dx.
And y is equal to this quotient from the given problem,
and then, we're going to multiply it by this fraction here.
So now, we need to simplify.
Notice how one factor of x is going to
simplify out here and here.
So this will be x to the 3rd.
Then looking at the factor of x minus three,
this is x minus three to the 1/2.
This is x minus three to the 1st.
So this will simplify out and this will
become x minus three to the 1/2 power.
And then, notice how we have four factors
of x plus one in the denominator,
three here and one here.
So we'll have 3x cubed times our trinomial,
Our denominator is going to be two times
four factors of x plus one
and the quantity x minus three to the 1/2 power.
And so you can see, this one was quite a bit of work,
but this would be our derivative function
by performing logarithmic differentiation.
I hope this was helpful.