### 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, weβve been given
a multivariable function. Thatβs a function in terms of more
than one variable. Here, those are π₯, π¦, and π§. Weβre looking to find the first
partial derivative of our function with respect to π¦. And we denote this as shown. Essentially, what weβre now doing
is seeing how the function changes as we let just one of our variables change. In this case, thatβs 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§, weβre going to treat π₯
and π§ both as constants. Then, weβll 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 weβre 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, weβre simply left with
two π§. And so we found the first partial
derivative with respect to π¦ of our function. Itβs π₯ cubed π§ squared plus two
π§.

And of course, had we chosen to do
so, we couldβve found the first partial derivative with respect to π₯ and with
respect to π§. Letβs see what that mightβve 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 weβre 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.

Letβs 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, weβre
treating the π¦ as a constant. So weβre left with two π¦. And the first partial derivative
with respect to π§ is two π₯ cubed π¦π§ plus two π¦.