# Video: Finding the First Partial Derivative of a Three-Variable Function

Find the first partial derivative with respect to π¦ of π(π₯, π¦, π§) = π₯Β³π¦π§Β² + 2π¦π§.

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### Video Transcript

Find the first partial derivative with respect to π¦ of π of π₯, π¦, π§ equals π₯ cubed π¦π§ squared plus two π¦π§.

In this question, weβve been given a multivariable function. Thatβs a function in terms of more than one variable. Here, those are π₯, π¦, and π§. Weβre looking to find the first partial derivative of our function with respect to π¦. And we denote this as shown. Essentially, what weβre now doing is seeing how the function changes as we let just one of our variables change. In this case, thatβs π¦. And to do so, we hold all the other variables constant.

And so, when we find the first partial derivative of π₯ cubed π¦π§ squared plus two π¦π§, weβre going to treat π₯ and π§ both as constants. Then, weβll perform this process term by term. It might be helpful to write π₯ cubed π¦π§ squared as π₯ cubed π§ squared π¦. And then, treating π₯ cubed π§ squared as a constant and differentiating with respect to π¦, we see weβre just left with the constant part, with π₯ cubed π§ squared. Similarly, when we differentiate two π¦π§ with respect to π¦ and we treat π§ as a constant, weβre simply left with two π§. And so we found the first partial derivative with respect to π¦ of our function. Itβs π₯ cubed π§ squared plus two π§.

And of course, had we chosen to do so, we couldβve found the first partial derivative with respect to π₯ and with respect to π§. Letβs see what that mightβve looked like. When we differentiate π₯ cubed with respect to π₯, we get three π₯ squared. Now, we treat π¦π§ squared as a constant here, and we get three π₯ squared π¦π§ squared. Then, when weβre finding the first partial derivative with respect to π₯, we treat π¦ and π§ as a constant. So the derivative of two π¦π§ is zero. And then, the first partial derivative with respect to π₯ of our function is three π₯ squared π¦π§ squared.

Letβs perform this process one more time, this time finding the first partial derivative with respect to π§. Now, if we were to differentiate π§ squared with respect to π§, we just get two π§. So we treat π₯ cubed and π¦ like a constant, and we find that the first term becomes two π₯ cubed π¦π§. Then, in our second term, weβre treating the π¦ as a constant. So weβre left with two π¦. And the first partial derivative with respect to π§ is two π₯ cubed π¦π§ plus two π¦.