### Video Transcript

Given that π¦ is equal to three root π₯ minus two π₯ divided by root π₯, determine the derivative of π¦ with respect to π₯.

Weβre given that π¦ is some function of π₯. In fact, π¦ is the quotient of two functions. And we need to determine the derivative of π¦ with respect to π₯. And since π¦ is the quotient of two functions, we might be tempted to do this by using the quotient rule, and this would work. However, we can notice an easier method for doing this. We can divide both terms in our numerator by the square root of π₯ by using our laws of exponents. So, weβll start by rewriting π¦ as three root π₯ divided by root π₯ minus two π₯ divided by root π₯.

And now, we want to simplify this expression. In our first term, root π₯ divided by root π₯ is equal to one. To simplify our second term, we need to recall two of our laws of exponents. π₯ to the power of π divided by π₯ to the power π is equal to π₯ to the power of π minus π. And we also need to remember we can rewrite the square root of π₯ as π₯ to the power of one-half. So, by using these two rules and remembering that π₯ is equal to π₯ to the first power, we can divide π₯ by root π₯. We get that itβs equal to π₯ to the power of one-half, which is equal to root π₯.

So, weβve rewritten π¦ as three minus two root π₯, and we can differentiate this term by term by using the power rule for differentiation. And this is much simpler than using the quotient rule. We get dπ¦ by dπ₯ is equal to the derivative of three minus two root π₯ with respect to π₯. And it might be easier to write the square root of π₯ as π₯ to the power of one-half. We now want to differentiate this term by term by using the power rule for differentiation. We multiply by our exponent of π₯ and reduce this exponent by one.

In our first term, the derivative of the constant three is equal to zero. Then, to differentiate our second term, we use our power rule for differentiation. We multiply by the exponent of one-half and reduce this exponent by one. This gives us one-half times negative two multiplied by π₯ to the power of one-half minus one. And we can simplify this to get negative π₯ raised to the power of negative one-half. And we could leave our answer like this. However, weβll use our laws of exponents to rewrite π₯ to the power of negative one-half as one divided by the square root of π₯, and this gives us our final answer.

We were able to show if π¦ is equal to three root π₯ minus two π₯ all divided by the square root of π₯, then dπ¦ by dπ₯ is equal to negative one divided by the square root of π₯.