### Video Transcript

Given that π¦ is equal to π§ squared divided by three plus nine π§ squared and π§ is equal to the square root of eight π₯ plus three, determine dπ¦ by dπ₯.

We need to find dπ¦ by dπ₯. Thatβs the first derivative of π¦ with respect to π₯. But weβre not given π¦ as a function in π₯. Instead, weβre given π¦ as a function in π§, and we can see that π§ is also given as a function in π₯. This means weβll need to find dπ¦ by dπ₯ by using the chain rule. Recall, this tells us if π¦ is a function in π§ and π§ in turn is a function in π₯, then dπ¦ by dπ₯ will be equal to dπ¦ by dπ§ times dπ§ by dπ₯. So now we need to differentiate π¦ with respect to π§ and then π§ with respect to π₯.

Letβs start with differentiating π¦ with respect to π§. We can see this is the quotient of two polynomials. So weβll do this by using the quotient rule. So weβll start by recalling the quotient rule. If π’ and π£ are differentiable functions, then the derivative of π’ of π§ divided by 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. So to find dπ¦ by dπ§, weβll set π’ of π§ to be the function in our numerator, thatβs π§ squared, and π£ of π§ to be the function in our denominator, thatβs three plus nine π§ squared.

And to apply the quotient rule, weβre going to need to find expressions for π’ prime of π§ and π£ prime of π§. We can do both of these by using the power rule for differentiation. We see that π’ prime of π§ will be equal to two π§ and π£ prime of π§ will be equal to 18π§. Now that we found expressions for π’ prime and π£ prime, we can find dπ¦ by dπ§ by using the quotient rule. Substituting in our expressions for π’, π£, π’ prime, and π£ prime, we get two π§ times three plus nine π§ squared minus 18π§ times π§ squared all divided by three plus nine π§ squared all squared.

And we can simplify this expression. In our numerator, if we distribute over the parentheses, one of our terms will be 18π§ cubed. However, we also have in our numerator negative 18π§ cubed. So all thatβs left in our numerator is six π§. Therefore, dπ¦ by dπ§ is equal to six π§ divided by three plus nine π§ squared all squared.

So we found an expression for dπ¦ by dπ§ . All thatβs left now is to find dπ§ by dπ₯. So we want to find dπ§ by dπ₯, and π§ is equal to the square root of eight π₯ plus three. Thereβs a few different ways we could differentiate this. We could use the chain rule or we could use the general power rule. We can use either method; weβre going to use the general power rule. First, weβll rewrite π§ as eight π₯ plus three all raised to the power of one-half. Next, we need to recall the general power rule. This tells us for differentiable function π of π₯ and any 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.

In our case, our exponent π is one-half and our inner function π of π₯ is eight π₯ plus three. And if π of π₯ is the linear function eight π₯ plus three, then π prime of π₯ will be the slope of this function, which is eight. Weβre now ready to find an expression for dπ§ by dπ₯ by using the general power rule. Our value of π is one-half and π prime of π₯ is equal to eight. So we get dπ§ by dπ₯ is equal to one-half times eight multiplied by eight π₯ plus three all raised to the power of negative one-half. And we can simplify this. First, one-half multiplied by eight is equal to four. Next, eight π₯ plus three all raised to the power of negative one-half is the same as dividing by the square root of eight π₯ plus three.

So we found an expression for dπ§ by dπ₯. Itβs equal to four divided by the square root of eight π₯ plus three. So we found an expression for dπ¦ by dπ§ and dπ§ by dπ₯. This means we can use the chain rule to find an expression for dπ¦ by dπ₯. So by substituting in our expressions for dπ¦ by dπ§ and dπ§ by dπ₯, we get dπ¦ by dπ₯ is equal to four divided by the square root of eight π₯ plus three multiplied by six π§ divided by three plus nine π§ squared all squared. And at this point, we can simplify. Remember, in the question, weβre told that π§ is equal to the square root of eight π₯ plus three, so we could start substituting this into our expression.

However, notice in our numerator, we have π§ and in our denominator, we have the square root of eight π₯ plus three, which is equal to π§. So we can cancel this shared factor of π§ in our numerator and denominator. This gives us 24 divided by three plus nine π§ squared all squared. Now weβll use our substitution π§ is equal to the square root of eight π₯ plus three, which of course means that π§ squared will just be equal to eight π₯ plus three. So by using the fact that π§ squared is eight π₯ plus three, we get dπ¦ by dπ₯ is equal to 24 divided by three plus nine times eight π₯ plus three all squared.

And thereβs only a little bit more simplification we need to do. Weβll distribute nine over our parentheses. This gives us 72π₯ plus 27. And finally we have three plus 27 is equal to 30. And this is our final answer. Therefore, by using the chain rule, the quotient rule, and the general power rule, we were able to show if π¦ is equal to π§ squared over three plus nine π§ squared and π§ is equal to the square root of eight π₯ plus three, then dπ¦ by dπ₯ is equal to 24 divided by 72π₯ plus 30 all squared.