We are given f of x and asked to find
f prime of x using logarithmic differentiation.
Let's first rewrite the function and replace f of x with y.
This gives us y equals the product of x cubed
and the fifth power of x minus nine
divided by the sixth power of x squared plus seven.
And now we will take the natural log
of both sides of the equation.
And now we will use the properties of logarithms
to expand the right side of the equation.
Let's first deal with the quotient.
And since the natural log of x divided by y
is equal to natural log x minus natural log y,
on the right we will have the natural log of the numerator
minus the natural log of the denominator.
So, again, we have natural log y equals
natural log of the numerator,
which is x cubed times the fifth power of x minus nine.
And then minus the natural log of the denominator,
which is the sixth power of x squared plus seven.
And now we can expand natural log of x cubed
times the fifth power of x minus nine
by using the product property of logarithms
where the natural log of x times y
is equal to natural log x plus natural log y,
which means you can write natural log of x cubed
times the fifth power of x minus nine
as natural log of x cubed plus natural log
of the fifth power of x minus nine.
Then we can expand one more time by using
the power property of logarithms,
which is natural log of x to the power of y
is equal to y times natural log x.
So, applying the power property of logs
we have natural log y is equal to natural log of x cubed
is equal to three times natural log x
plus the natural log of the fifth power of x minus nine
is equal to five times natural log
of the quantity x minus nine.
And then we have minus natural log of the sixth power
of x squared plus seven is equal to six times natural log
of the quantity x squared plus seven.
And now for the next step we will differentiate
both sides of the equation with respect to x.
On the left the derivative of natural log y
with respect to x requires the chain rule.
The derivative is equal to one divided by y times dy/dx.
Again, we're differentiating with respect to x,
and because we have a y term we have this extra factor
of dy/dx due to the chain rule.
And then equals on the right,
the derivative of three natural log x with respect to x
is equal to three times the derivative of natural log x,
which is one divided by x.
And then plus the derivative of five times
natural log of the quantity x minus nine.
If we're not sure if we need to apply the chain rule
we are never wrong to apply it.
Notice in this case the inner function u is x minus nine,
and since u prime is equal to one
the chain rule is not required.
The derivative is equal to five times
one divided by the quantity x minus nine,
and then we have minus the derivative
of six times natural log of the quantity
x squared plus seven.
Here we are required to use the chain rule
where u is equal to x squared plus seven.
The derivative is equal to six times
the derivative of natural log
of the quantity x squared plus seven,
which is equal to one divided by the quantity
x squared plus seven times u prime,
which is the derivative of x squared plus seven
with respect to x, which is two x.
Let's go and simplify these products.
We have one divided by y times dy/dx
equals three divided by x plus five
divided by the quantity x minus nine minus,
in the numerator here we have 12x
divided by the quantity x squared plus seven.
The next step is to solve the equation for dy/dx
by multiplying both sides of the equation by y.
Let's do this on the next slide.
So, again, now to solve for dy/dx we will multiply
both sides of the equation by y.
On the left y times one divided by y simplifies to one
leaving us with dy/dx equals,
on the right we want to have the derivative in terms of x,
not x and y.
So, now we'll replace y with the original function,
which is the product of x cubed
and the fifth power of x minus nine
divided by the sixth power of x squared plus seven.
And then of course we have times three divided by x
plus five divided by the quantity x minus nine
minus 12x divided by the quantity x squared plus seven.
So, we probably could leave the derivative
in this form here,
but let's go ahead and clear
the parentheses by distributing.
We have dy/dx equals,
for this first product notice one factor of x
will simplify out between the numerator and denominator,
and therefore the product is going to be three x squared
times the fifth power of x minus nine
all divided by the sixth power of x squared plus seven.
And then notice for the second product,
one factor of x minus nine,
between the numerator and denominator will simplify out
leaving us with plus five x cubed times the quantity
x minus nine to the fourth divided by the sixth power
of x squared plus seven.
And then minus for the last product
nothing will simplify out,
multiplying we have 12x to the fourth
times the fifth power of x minus nine
all divided by the seventh power of x squared plus seven.
And now looking at the fraction notice how
the first two fractions do have the same denominator
so let's go ahead and combine those two fractions.
We have dy/dx, or we'll call it f prime of x now
because that's what the question asked for,
is equal to, again, combining the first two fractions
we know the denominator is going to be the sixth power
of x squared plus seven.
And the numerator is going to be three x squared
times the fifth power of x minus nine
plus five x cubed times the fourth power of x minus nine,
and then we have minus 12x to the fourth
times the quantity x minus nine to the fifth
all divided by the seventh power of x squared plus seven.
And we'll go ahead and leave the derivative function
in this form here.
But again, we probably would've been fine
by leaving the derivative in this form here.
I hope you found this helpful.