### Video Transcript

Find the first partial derivative with respect to π¦ of π of π₯, π¦ equals π₯ squared π¦ minus three π¦ to the fourth power.

Weβve been given a multivariable function. Thatβs a function in terms of both π₯ and π¦. And weβre being asked to find the first partial derivative with respect to π¦ of the function. We use curly dβs as shown to denote this. Now, when a function is made up of multiple variables, we look to see how the function changes as we let just one of those variables change. And we hold all the others constant.

In this case, weβre going to let π¦ change. And so we treat the other variable π₯ as a constant. Aside from this, the rules for differentiation are roughly the same. Letβs go term by term. Letβs differentiate π₯ squared π¦.

Since weβre imagining π₯ as a constant, we can also imagine π₯ squared to be a constant. And when we differentiate a constant times π¦ with respect to π¦, weβre just left with that constant. And so the first part of our partial derivative is just π₯ squared. Next, we differentiate negative three π¦ to the fourth power. This is simply a power term. So we differentiate as normal. We multiply the entire term by the exponent and then reduce that exponent by one. So itβs negative four times three π¦ cubed. This partial derivative simplifies to π₯ squared minus 12π¦ cubed. And so we found the first partial derivative with respect to π¦ of our function.

And of course, we could have done this with respect to π₯. This time, though, we would be treating π¦ as a constant. So the derivative of π₯ squared with respect to π₯ is two π₯. And that means if we treat π¦ as a constant, the first part of our first partial derivative with respect to π₯ of our function is two π₯π¦. Then, since we would be treating π¦ as a constant, π¦ to the fourth power, and therefore negative three π¦ to the fourth power, is also a constant. And we know when we differentiate a constant with respect π₯, we get zero. And so the first partial derivative with respect to π₯ of our function would have been two π₯π¦ had that been required.