Find F prime of x for the given function
using log differentiation.
For the first step,
let's replace f of x with y,
and write the cube root using
the irrational x power of one third.
The given function is equivalent to y
equals,
the quantity x cubed plus five x divided
by the quantity x to the six plus three
raised to the one third power.
And now to apply log differentiation,
we take the natural log of both sides
of the equation
and now we expand the right side as much
as possible because we have
a quotient we can use the quotient property
of log rhythms where the log of m divided by n
is equal to log m minus log n
which means you can write the right side
as natural log of x to the third plus
five x minus natural log
of the quantity x to the six plus three
raised to the one third power.
And now we can apply the power property of log rhythms
to the second log here,
we can move the exponent of
one third to the position of the coefficient.
Which gives us natural log y
equals natural log of the quantity x to the third
plus five x
minus one third
times a natural log of the quantity
x to the six plus three.
The next step is to differentiate
both sides of the equation
with respect to x.
The derivative of natural log y
with respect to x
is going to require the chain rule,
reverse differentiate with respect to y
which gives us one over y
now we multiply it by a factor
of d y d x.
This is implicit differentiation,
equals on the right differentiating
with respect to x,
the derivative of natural log of the quantity
x to the third
plus five x
is equal to one over times u prime,
or one over the quantity x to the third plus five x,
times the derivative of x to the third
plus five x, which is
three x squared plus five.
Now we have minus one third
times the derivative
of natural log x to the six plus three
with respect to x.
Which again is one over u times u prime,
or one over x to the six plus three
times six x to the fifth.
Lets go ahead and simplify
on the right, for the first product
nothing simplifies
we have the quantity three x squared
plus five
over the quantity x to the third
plus five x
minus, for the second product we have
a common factor of three between,
six and three
three simplifies to one
six simplifies to two,
which gives us minus two x to the fifth
all over
x to the sixth plus three.
And now when you dissolve four d y d x
by multiplying,
both sides of the equation by y
On the left,
y times one over y simplifies
to one one times d y d x is
d y d x
we have d y d x equals y times the difference
of the irrational expressions.
So while this is the derivative,
we prefer to express the derivative
with respect to x only in terms of x.
So now we replace y
with the original function.
Lets do this on the next slide.
So now we do have d y d x only in terms of x,
so we could leave it in this form.
But lets go ahead and distribute this first fraction,
and then simplify as much as possible.
And I'm going to go ahead and put all the numerators
and denominators in parentheses,
to make the simplifying more clear.
Looking at this first product
notice how x q plus five x divided by itself,
simplifies to one,
giving us d y d x,
is equal to
for this first product we have
three x squared plus five,
all over the quantity x to the sixth plus three
to the one third power.
Now looking at the second product,
nothing simplifies,
but when multiplying the denominators,
because the bases are the same,
we add the exponents,
one third plus one is four thirds,
so the denominators going to be
the quantity x to the sixth
plus three to the four thirds power.
In the numerator we distribute
two x to the fifth,
which gives us two x to the eighth,
and then plus ten x to the sixth.
So while you may be able to leave the
derivative in this form here,
it may also be expressed in this final form here.
I hope you found this helpful