"Derivative with respect to..." (partial derivatives)

Added: 2026-05-23

[00:00:00]Let's examine these form of questions where you have a derivative with the respect to something. What is that respect to something? In these particular cases, you are looking here at a function, let's say Z is equal to something with respect to two independent variables X and Y.

[00:00:15]I have here a representation Z is equal to X + Y over Y. This here is a representation. You can graph this on the three-dimensional software. You'll have here on an X Y and Z axis a representation. You can see what it looks like. But this is what I have X Y Z axis. X and Y are independent variables. Now let's come here derivative with respect to my X variable. Now I'm going to do the derivative of this. I can represent it as this. I'm doing this partial derivative representation where the X will be differentiated, the Y will be treated as a constant or coefficient.

[00:00:48]I can push the Y out. Here I'm really looking at 1 over Y, then I can say I'm doing partial derivative with respect to your X variable of what? Of this item right here, X + Y. And then you can take this all the way through. Derivative of this Y, which is a constant in this particular case, is zero. Derivative of X is a one. You're left here with your end result, which will be 1 over Y. Your derivative here end result here is good.

[00:01:11]You're doing here with respect to the X variable, and your answer 1 over Y is correct. Now let's do something here with respect to the Y variable, and I'll show you what that will be. In this particular case with the same function right here, Z is equal to X + Y over Y, let's do derivative with respect to Y.

[00:01:26]In that particular case, you're looking here at this representation, and you have this. Now it doesn't make any sense to push the Y out cuz here your Y is a variable, your X is not. You have to do the quotient rule over here. You can do Y times the partial derivative with respect to the Y variable of X + Y minus X + Y times the partial derivative here with respect to the Y variable of your denominator, which will be Y, over your denominator Y squared. This is my quotient rule. Y times here X here is going to be a zero, Y is a one, you'll have a Y.

[00:02:00]Here you have X plus Y over Y squared. Simplify this out, Y minus Y cancel out, you'll end up here with minus X over Y squared. And this right here is your end result.

[00:02:14]The partial derivative with respect to X, one over Y. Partial derivative with respect to Y, minus X over Y squared.

[00:02:21]Both of these end results are correct.