### Video Transcript

Find dπ¦ by dπ₯ at π₯ is equal to two when π¦ is equal to π₯ cubed divided by two π₯ minus three all raised to the fourth power.

Weβre given π¦ as a function in π₯, and weβre asked to determine dπ¦ by dπ₯ when π₯ is equal to two. So, the first thing weβre going to need to do is differentiate π¦ with respect to π₯. And we can see that π¦ is the quotient of two functions. So, weβll do this by using the quotient rule. Letβs start by recalling the quotient rule.

The quotient rule tells us for differentiable functions π’ of π₯ and π£ of π₯, the derivative of π’ of π₯ over π£ of π₯ with respect to π₯ is equal to π’ prime of π₯ times π£ of π₯ minus π£ prime of π₯ times π’ of π₯ all divided by π£ of π₯ all squared. And of course, we know this wonβt be valid when π£ of π₯ is equal to zero. So, weβll set π’ of π₯ to be the function in our numerator, thatβs π₯ cubed, and π£ of π₯ to be the function in our denominator. Thatβs two π₯ minus three all raised to the fourth power.

And we can see that both π’ of π₯ and π£ of π₯ are polynomials, so we can differentiate this by using the quotient rule. To use the quotient rule, weβre going to need to find the expressions for π’ prime of π₯ and π£ prime of π₯. Letβs start with π’ prime of π₯. Thatβs the derivative of π₯ cubed with respect to π₯. Well, we can do this by using the power rule for differentiation. We want to multiply by our exponent of π₯ and reduce this exponent by one. This gives us π’ prime of π₯ is equal to three π₯ squared.

Next, weβre going to want to find an expression for π£ prime of π₯. However, we can see this is the composition of two functions. Weβre raising a linear function to the fourth power. So, we have a few options to do this. We could use the chain rule or we could distribute this by using the binomial expansion formula. Weβll find π£ prime of π₯ by using the general power rule, which is just another example of the chain rule. We recall this tells us for differentiable function π of π₯ and constant π, the derivative of π of π₯ all raised to the πth power with respect to π₯ is equal to π times π prime of π₯ multiplied by π of π₯ raised to the power of π minus one.

To use this on our function π£ of π₯, weβll set π of π₯ to be our inner function, thatβs two π₯ minus three, and π to be our exponent, which is four. Then, to use the general power rule, we need to find an expression for π prime of π₯. Thatβs the derivative of two π₯ minus three with respect to π₯, which is equal to two. So, by using the general power rule with π equal to two, π prime of π₯ equal to two, and π of π₯ equal to two π₯ minus three, we get π£ prime of π₯ is equal to four times two multiplied by two π₯ minus three all cubed, which, of course, simplifies to give us eight times two π₯ minus three all cubed.

And now that we found the expressions for π’ prime of π₯ and π£ prime of π₯, we can apply the quotient rule to our original function. Substituting our expressions for π’ of π₯, π£ of π₯, π’ prime of π₯, and π£ prime of π₯ into our quotient rule formula, we get dπ¦ by dπ₯ is equal to three π₯ squared times two π₯ minus three all raised to the fourth power minus eight times two π₯ minus three all cubed times π₯ cubed all divided by two π₯ minus three all raised to the fourth power all squared. And weβll simplify this equation slightly. First, we have two π₯ minus three all raised to the fourth power all squared. By using our laws of exponents, we can just write this as two π₯ minus three all raised to the eighth power. And we could simplify this even further.

However, remember, we only need to find the derivative when π₯ is equal to two. So weβre going to substitute π₯ is equal to two into this expression. Substituting π₯ is equal to two into our expression for dπ¦ by dπ₯ gives us the following expression. And all we need to do now is evaluate this expression. We could do this by using a calculator. However, this is not necessary.

First, we need to notice that two times two minus three is equal to one. So, all three of these factors just simplify to give us one. And, of course, one to the fourth power, one cubed, and one to the eighth power are all equal to one. So, this entire expression simplifies to give us three times two squared minus eight times two cubed. And we can just calculate this; itβs equal to negative 52.

Therefore, by using the quotient rule, we were able to show if π¦ is equal to π₯ cubed over two π₯ minus three all raised to the fourth power, then dπ¦ by dπ₯ when π₯ is equal to two is given by negative 52.