### Video Transcript

Differentiate π¦ is equal to π₯ minus one times π₯ plus one times π₯ squared plus one all divided by π₯.

We can see the question is asking us to differentiate the quotient to two functions. And we can do this by using the quotient rule for differentiation, which tells us for functions π’ and π£ the derivative of π’ divided by π£ is equal to π£π’ prime minus π’π£ prime all divided by π£ squared. So to use the quotient rule, weβll set π’ as the function in our numerator. Thatβs π₯ minus one times π₯ plus one times π₯ squared plus one. And weβll set π£ to be the function in our denominator. Thatβs π£ is equal to π₯.

We now need to find expressions for π’ prime and π£ prime. We can see that π’ is a product to function. So we might be tempted to do this by using the product rule. However, this is not necessary since we can simplify our expression for π’ by using a difference between squares.

We know that π squared minus π squared is equal to π minus π times π plus π. This tells us that π₯ minus one multiplied by π₯ plus one is equal to π₯ squared minus one. So this gives us π’ is equal to π₯ squared minus one times π₯ squared plus one. But then we can use a difference between squares again. Doing this, we get π’ is equal to π₯ to the fourth power minus one.

Weβre now ready to find expressions for π’ prime and π£ prime. Weβll start with π’ prime. Thatβs the derivative of π₯ to the fourth power minus one with respect to π₯. And we can evaluate this derivative by using the power rule for differentiation, which 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 and reduce the exponent by one.

Remember, we can differentiate each term separately. Letβs start by differentiating π₯ to the fourth power with respect to π₯. We multiply by our exponent of four and then reduce this exponent by one. This gives us four π₯ to the power of four minus one. And of course, we can simplify four minus one to be three.

Now we need to differentiate negative one. We could do this by writing negative one as negative π₯ to the zeroth power. However, we could also do this by remembering the derivative of any constant is equal to zero. This means weβve shown that π’ prime is equal to four π₯ cubed.

We can do something similar to find π£ prime. Thatβs the derivative with respect to π₯ of π₯. Weβll do this by rewriting π₯ as π₯ to the first power. We then differentiate this by using the power rule for differentiation. We multiply by our exponent of one and then reduce this exponent by one. This gives us one times π₯ to the power of one minus one. We can simplify one minus one to give us zero. But π₯ to the zeroth power is just equal to one. So weβve shown that π£ prime is just equal to one.

Weβre now ready to find the derivative of π¦. First, by the quotient rule, we have that this is equal to π£π’ prime minus π’π£ prime divided by π£ squared. Next, we substitute our expressions for π’, π£, π’ prime, and π£ prime. This gives us π₯ times four π₯ cubed minus π₯ to the fourth power minus one times one all divided by π₯ squared.

We can now simplify this expression. We have π₯ times four π₯ cubed is equal to four π₯ to the fourth power. Next, multiplying by one doesnβt change anything. And then negative one times π₯ to the fourth power minus one is equal to negative π₯ to the fourth power plus one. Finally, we divide this through by π₯ squared.

And now we can simplify this further. We have four π₯ to the fourth power minus π₯ to the fourth power is equal to three π₯ to the fourth power. This gives us three π₯ to the fourth power plus one all divided by π₯ squared. And we could leave our answer like this. However, weβre going to divide through by π₯ squared. Three π₯ to the fourth power divided by π₯ squared is three π₯ squared. And then we add one over π₯ squared, which by our laws of exponents is equal to π₯ to the power of negative two.

Therefore, weβve shown the derivative of π₯ minus one times π₯ plus one times π₯ squared plus one all divided by π₯ with respect to π₯ is equal to three π₯ squared plus π₯ to the power of negative two.