### Video Transcript

Find the first partial derivative with respect to π¦ of the function π of π₯, π¦ equals π₯ squared minus π¦ squared plus six π₯π¦ plus four π₯ minus eight π¦ plus two.

Here, we have a multivariable function with two variables. Itβs a function in terms of both π₯ and π¦. Weβre being asked to find the first partial derivative with respect to π¦ of our function. Thatβs denoted as shown, and itβs sometimes pronounced as ππ ππ¦. But what does it mean to find the partial derivative of a multivariable function?

Well, weβre looking to see what the behavior of the function is as we let just one of the variables change and we hold all the others constant. In this case, weβre finding the first partial derivative with respect to π¦. So, weβre going to treat π₯ as a constant. Aside from this, our usual rules for differentiation apply. Letβs look at this term by term.

We begin by looking at the first term, π₯ squared. Now, we said weβre going to treat π₯ as a constant. So, π₯ squared itself we can also imagine as a constant, and the derivative of a constant is zero. Then, we move on to our second term, which is negative π¦ squared. We differentiate as we would any other power term by multiplying the entire term by the exponent and reducing that exponent by one. So, the derivative of negative π¦ squared with respect to π¦ is negative two π¦.

Next, our third term, once again, we treat π₯ as a constant, which means that we can imagine six π₯ itself is also a constant. Now, we know that the derivative with respect to π¦ of some constant times π¦ is simply that constant. So, the derivative of six π₯π¦ becomes just six π₯.

Then, four π₯ we treat it as a constant, and we know that the derivative of a constant is zero. Then, we look at our fifth term. The derivative of negative eight π¦ with respect to π¦ is negative eight. We have one final constant, and the derivative of two with respect to π¦ is zero.

So, we can say that the first partial derivative with respect to π¦ of our function is negative two π¦ plus six π₯ minus eight. Since each term shares a factor of two, it makes sense to factor our expression. And so, we found the first partial derivative with respect to π¦ of our function. Itβs two times three π₯ minus π¦ minus four.