# Question Video: Finding the First Partial Derivative of a Multivariable Function with Two Variables

Find the first partial derivative with respect to π¦ of the function π(π₯, π¦) = π₯Β² β π¦Β² + 6π₯π¦ + 4π₯ β 8π¦ + 2.

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### 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. Itβs a function in terms of both π₯ and π¦. Weβre being asked to find the first partial derivative with respect to π¦ of our function. Thatβs denoted as shown, and itβs sometimes pronounced as ππ ππ¦. But what does it mean to find the partial derivative of a multivariable function?

Well, weβre 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, weβre finding the first partial derivative with respect to π¦. So, weβre going to treat π₯ as a constant. Aside from this, our usual rules for differentiation apply. Letβs look at this term by term.

We begin by looking at the first term, π₯ squared. Now, we said weβre 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. Itβs two times three π₯ minus π¦ minus four.