### Video Transcript

Differentiate π¦ is equal to two π₯
to the seventh power minus six π₯ to the fifth power plus four π₯ cubed minus eight
all divided by four π₯ to the fifth power.

The question wants us to find the
derivative of the quotient of two polynomials. And since the question wants us to
differentiate the quotient of two functions, weβll do this by using the quotient
rule. We recall the quotient rule tells
us the derivative of the quotient of two functions π’ over π£ is equal to π£ times
π’ prime minus π’ times π£ prime all divided by π£ squared.

So, to apply the quotient rule,
weβll set π’ of π₯ to be the polynomial in our numerator. Thatβs two π₯ to the seventh power
minus six π₯ to the fifth power plus four π₯ cubed minus eight. And weβll set π£ of π₯ to be the
polynomial in our denominator. Thatβs four π₯ to the fifth
power. Now, to apply the quotient rule,
weβre going to need to find expressions for π’ prime of π₯ and π£ prime of π₯. Since both of these are
polynomials, we can do this term by term by using the power rule for
differentiation.

We recall this tells us for
constants π and π, the derivative of ππ₯ to the πth power with respect to π₯ is
equal to π times π times π₯ to the power of π minus one. We multiply by the exponent of π₯
and then reduce this exponent by one. Applying this rule to each term in
π’ of π₯, we get π’ prime of π₯ is equal to 14π₯ to the sixth power minus 30π₯ to
the fourth power plus 12π₯ squared. And we can do the same to find an
expression for π£ prime of π₯, we get 20π₯ to the fourth power.

Weβre now ready to find an
expression for the derivative of π¦ with respect to π₯ by using the quotient
rule. We get that itβs equal to π£ of π₯
times π’ prime of π₯ minus π£ prime of π₯ times π’ of π₯ all divided by π£ of π₯
squared. Substituting in our expressions for
π£ of π₯, π’ of π₯, π’ prime of π₯, and π£ prime of π₯. We get that π¦ prime of π₯ is equal
to four π₯ to the fifth power times 14π₯ to the sixth power minus 30π₯ to the fourth
power plus 12π₯ squared minus 20π₯ to the fourth power multiplied by two π₯ to the seventh
power minus six π₯ to the fifth power plus four π₯ cubed minus eight all divided by
four π₯ to the fifth power all squared.

It might be tempting to start
distributing over our parentheses. However, itβs easier to simplify
this expression by factoring. Initially, weβll simplify our
denominator. We have four π₯ to the fifth power
all squared is equal to 16π₯ to the power of 10. We could now start simplifying. First, our numerator and our
denominator share a factor of four π₯ to the fourth power. So, letβs divide our numerator and
our denominator by four π₯ to the fourth power.

First, dividing the first term in
our numerator by four π₯ to the fourth power, weβre left with π₯. Next, to divide the second term in
our numerator by four π₯ to the fourth power, weβll simplify the first factor to
just be five. Finally, we need to divide our
denominator by four π₯ to the fourth power. We see this leaves us with four π₯
to the sixth power, and this gives us the following expression. At this point, thereβs several
different methods we could use to simplify this expression even further. We could continue factoring our
numerator; however, weβll simplify this by distributing over our parentheses and our
numerator.

So, letβs start by distributing π₯
over the parentheses in our first term. We need to multiply each term
inside of our parentheses by π₯. And to do this, we just need to add
one to each of our exponents of π₯. This gives us 14π₯ to the seventh
power minus 30π₯ to the fifth power plus 12π₯ cubed. We now want to do the same with the
second term in our numerator. However, remember, this time, there
are four terms inside of our parentheses. And we need to multiply each term
by negative five. Doing this and being careful with
our signs, we get negative 10π₯ to the seventh power plus 30π₯ to the fifth power
minus 20π₯ cubed plus 40.

Finally, remember, we need to
divide all of this by four π₯ to the sixth power. We can now simplify our numerator
by combining like terms. First, negative 30π₯ to the fifth
power plus 30π₯ to the fifth power is equal to zero. Next, if we have 14π₯ to the
seventh power and we subtract 10π₯ to the seventh power, weβre just left with four
π₯ to the seventh power. Similarly, if we have 12π₯ cubed
and then we subtract 20π₯ cubed, weβll be left with negative eight π₯ cubed. So, weβve now simplified this
expression to give us four π₯ to the seventh power minus eight π₯ cubed plus 40 all
divided by four π₯ to the sixth power.

And we could leave our answer like
this. However, weβll divide each term in
our numerator through by our denominator. So, letβs do this term by term. First, to divide four π₯ to the
seventh power by four π₯ to the sixth power, we can cancel the shared factor of four
and we can cancel six of the shared factors of π₯. This just leaves us with π₯. Next, in our second term, we have
eight divided by four will leave us with two and then π₯ cubed divided by π₯ to the
sixth power will leave us with π₯ cubed in our denominator. So, our second term simplify to
give us negative two divided by π₯ cubed.

Finally, in our third term, we have
40 divided by four is equal to 10. So, our third term simplifies to
give us 10 divided by π₯ to the sixth power. And again, we could leave our
answer like this. However, weβll use our laws of
exponents to rewrite our divisions as multiplications. So, by using the fact that dividing
by π₯ cubed is the same as multiplying by π₯ to the power of negative three and
dividing by π₯ to the sixth power is the same as multiplying by π₯ to the power of
negative six. We were able to rewrite our answer
as π₯ minus two π₯ to the power of negative three plus 10π₯ to the power of negative
six. And this is our final answer.

Therefore, we were able to show by
using the quotient rule. If π¦ is equal to two π₯ to the
seventh power minus six π₯ to the fifth power plus four π₯ cubed minus eight all
divided by four π₯ to the fifth power. Then the derivative of π¦ with
respect to π₯ is equal to π₯ minus two π₯ to the power of negative three plus 10π₯
to the power of negative six.