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# Implicit differentiation (advanced example)

## Video transcript

so let's take another implicit derivative of the somewhat crazy relationship and I have graphed the relationship here as you can see it is actually quite bizarre e to the x times y squared is equal to X minus y this is a sampling the range that's depicted here the X's and the Y's that satisfy this relationship well let's take the derivative of both sides so we'll apply our derivative operator to both sides and actually this is a good chance to maybe explore some different notation that's the we tend to be using that notation but often times you will see a derivative operator that just looks like a big capital D so maybe we'll do that right over here so let me make it clear this is this is the equivalent of saying D over DX I'll just use this big capital D operator so that you are familiar with the note with the notation and instead of using dy DX for the derivative of Y with respect to X I'm just going to write that as Y prime so we get a little bit of practice with different notation so let's take the derivative of this thing right over here well we're going to apply the chain rule actually going to apply the chain rule multiple times here the derivative of e to the something with respect to that something is going to be e to the something it's going to be e to the something times the derivative times the derivative of that something with respect to X so times the derivative of X Y squared so that's our left-hand side we are done taking the derivative yet and on our right hand side the derivative of X is just 1 and the derivative with respect to X of Y is just going to be minus or it's just write negative dy/dx but instead of writing dy/dx I'm going to write Y prime as you can see as you can tell I like that I like this notation and this notation more because it makes it explicit that I'm taking the derivative with respect to X here we just have to assume that we're taking the derivative with respect to X here we have to assume that's the derivative of Y with respect to X but anyway let's stick with this note a right over here and every time actually let me make all my let me make all of my Y Prime's all of my derivatives with respect to X all my derivatives of Y with respect to X let me make them pink so I can keep track of them so once again this is going to be equal to e to the X Y squared times the derivative of this well the derivative of this we can just use the product and actually a little bit of the chain rule here so the derivative of X is just 1 times the second function so it's going to be times y squared and then to that we're going to we're going to add the product of the first function which is this x times the derivative of Y squared with respect to X well that's going to be the derivative of Y squared with respect to Y which is just going to be 2y times the derivative of Y with respect to X which we are now writing as which we are now writing as Y prime Y prime and then that's going to be equal to 1 minus y prime that's going to be equal to 1 minus y prime and like we've been doing we now have to just solve for y prime so let's distribute this exponential this e to the X Y squared and we get we get e or maybe I should say Y squared Y squared times e to the X Y squared so that's that plus X or administrate to X Y e to the X Y squared Y prime the derivative of Y with respect to X is equal to is equal to 1 minus the derivative of Y with respect to X now let's get all of our Y Prime's on one side so let's add Y prime to both sides so let's add and just to be clear I'm adding 1 Y prime to both sides so let's add a Y prime to both sides and let's subtract this business from both sides so let's subtract Y squared e to the X Y squared to both sides by subtracting it from both sides so we're going to subtract y squared e to the X Y squared and we are left with we are left with 2x 2x Y e to the X Y squared plus 1 times y prime times y Prime we had this many Y Prime's and then we add another one Y prime so we have this many plus 1 Y Prime's that's going to be equal to that's going well I purposely added Y prime to both sides and so we are left with 1 1 minus is kind of a crazy expression y squared times e to the X y squared and now we just have to draw it divide both sides by this and we're left with the derivative of Y with respect to X is equal to this which I will just copy and paste actually let me just rewrite it it's equal to 1 over scroll down scroll down a little bit it's equal to 1 minus y squared e to the X Y squared over this business to let me give some more space to XY e to the X Y squared plus 1 and we're done it was kind of crazy but in fundamentally no different than what we've been doing in the last few examples