We want to find the first and second derivative
for the given function.
We will be applying the quotient role,
which is shown here in the upper right-hand corner.
F prime of X, is equal to,
let's begin with the denominator.
The denominator of the derivative
is the square of the denominator of the original function.
So if we square the quantity, one minus X squared,
we multiply the exponents
and we have four factors of one minus X.
And now for the numerator, we have the denominator times,
the derivative of the numerator.
Well, the nominator is the quantity one minus X squared.
And then times the derivative two of two X
with respect to X, which is two.
And then we have minus the numerator of two X,
times the derivative of the denominator,
to find the derivative of the square of one minus X,
we need to apply the chain rule.
We first multiply by two
and then subtract one from the exponent,
which gives us the quantity one minus X to the first,
and then times the derivative of the inner function.
The derivative of one minus X is negative one.
And of course we can drop the exponent
of one on the quantity, one minus X.
And now we need to simplify.
Notice how the two products in the numerator
have a common factor of one minus X.
Here we have two factors of one minus X,
and here we have one factor of one minus X.
So let's go ahead and factor out one factor
of the quantity one minus X from both products.
For the first product that leaves us
with one factor of one minus X,
times two.
In the second product, we're factoring out the one minus X,
which leaves us
with minus two X times two times negative one,
which is negative four X
and minus negative four X is the equivalent
of two plus four X.
Continuing to simplify.
Again, we have one factor of X.
We have one factor of one minus X in the numerator.
And four in the denominator,
the X minus one in the numerator
simplifies to one.
And the four factors of one minus X
simplifies to three factors of one minus X.
So the first derivative has a denominator
of the cube of one minus X.
And now in the numerator, if we distribute two,
we have two minus two X,
and then we still have. plus four X.
The numerator simplifies to two plus two X
or two X plus two.
And now we need to find the second derivative,
which is the derivative of the first derivative.
The denominator of the second derivative
is the square of the cube of one minus X,
which gives us six factors of one minus X
or the quantity one minus X to the sixth power.
In the numerator, we have the denominator
of the cube of one minus X,
times the derivative of the numerator.
The derivative of two X plus two is two.
And then we have minus the numerator of the quantity
two X plus two, times the derivative of the denominator,
which is going to be three
times the quantity one minus X squared
times the derivative of one minus X,
which is negative one.
And now again, we simplify.
Looking at the products in the numerator,
notice as the common factor of the square
of one minus X.
Let's factor out the square of one minus X
from both products.
In the first product that leaves us
with one factor of one minus X times two.
And then for the second product,
we're factoring out both factors of one minus X
so that leaves us with minus the quantity
two X plus two, times three, times negative one.
Subtracting a negative is equivalent to adding a positive.
So let's write this as plus three,
times the quantity two X plus two,
and this is still all of our six factors of one minus X.
Now simplify the factors of one minus X
in the numerator and denominator.
The two factors of one minus X and the numerator simplify
to one, and the six factors of one minus X
simplify to four factors of one minus X.
In the numerator if we distribute two,
we have two minus two X, and then we distribute three,
which gives us plus six X plus six.
Combining like terms, we have four X plus eight.
So the second derivative
is equal to the quantity four X plus eight,
divided by the quantity
one minus X raised to the fourth.
Now we could factor the four from four X plus eight,
but it's not going to simplify with the denominator,
so I'll go ahead and leave it in this form.
So here we have the first derivative
and here we have the second derivative.
I hope you found this helpful.