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# Implicit differentiation

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.D (LO)
,
FUN‑3.D.1 (EK)

## Video transcript

so we've got the equation x squared plus y squared is equal to 1 I guess we could call it a relationship and if we were to graph all of the points x and y that satisfy this relationship we get a unit circle like this and what I'm curious about in this video is how we can figure out the slope of the tangent line at any point of this unit circle and what immediately might be jumping out in your brain is well a circle defined this way this isn't a function it's not Y explicitly defined as a function of X for any x value you actually have two possible Y's that satisfy this relationship right over here so you might be tempted to maybe split this up into two separate functions of X you could say Y is equal to the positive square root of 1 minus x squared and you could say Y is equal to the negative square root of 1 minus x squared take the derivatives of each of these separately and you would be able to find the derivative for any X at or the derivative of the slope of the tangent line at any point but what I want to do in this video is literally leverage the chain rule to take the derivative implicitly so that I don't have to explicitly define Y is a function of X either way and the way we do that is literally just apply the derivative operator to both sides of this equation and then apply what we know about the chain rule because we are not explicitly defining Y as a function of X and explicitly getting Y is equal to F prime of X they call this which is really just a application of the chain rule we call it implicit differentiation implicit differentiation and what I want you to keep in the back of your mind the entire time is that it's just an application of the chain rule so let's apply the derivative operator to both sides of this so it's the derivative with respect to X of x squared plus y squared x squared plus y squared on the left hand side of our equation and then that's going to be equal to the derivative with respect to X on the right hand side I'm just doing the same exact thing to both sides of both sides of this equation now if I take the derivative of the sum of two terms that's the same thing as taking the sum of the derivative so this is going to be the same thing as the derivative with respect to X of x squared plus the derivative with respect to X of Y squared I'm writing all my orange stuff first so let's see this is going to be x squared it's going to be Y squared and then this is going to be equal to the derivative with respect to X of a constant this isn't changing with respect to X so we just get zero now this first term right over here we have done many many many many many times the derivative with respect to X of x squared is just the power rule here it's going to be two times X to the first power or we could just say 2x now what's interesting is what we're doing right over here the derivative with respect to X of Y squared and the realization here is to just apply the chain rule if we're taking the derivative with respect to X of this something we just have to take the derivative let me make it clear we're just going to take the derivative of our something the derivative of Y squared that's what we're taking that you can kind of view that as a function with respect to Y with respect to Y and then multiply that times the derivative of Y the derivative of Y with respect to X we're assuming that Y does change with respect to X Y is not some type of a constant that we're writing just in abstract terms so we're taking the derivative of this whole thing with respect to Y once again just the chain rule and then we're taking the derivative of Y with respect to X you it might be a little bit clearer if you kind of thought of it as the derivative with respect to X of of Y as a function of X Y as a function of X this might be or Y is a function of x squared which is essentially another way of writing what we had here this might be a little bit clearer in terms of the chain rule the derivative of Y is the function of x squared with respect to Y of X so the derivative of something squared the derivative of something squared with respect to that something times the derivative of that something with respect to X this is just the chain rule I want to say it over and over again this is just this is just the chain rule so let's do that what do we get on the right hand side over here and I'll write it over here as well this would be equal to the derivative of Y squared with respect to Y is just going to be two times y 2 times y just an application of the chain rule and the derivative of Y with respect to X well we don't know what that is so we're just going to leave that as times the derivative of Y times the derivative of Y with respect to X so let's just write this down over here so what we have is 2x 2x plus 2x plus derivative of something squared with respect to that something is 2 times the something and Kate in this case the something is y so 2 times y and then times the derivative of Y the derivative of Y with respect to X and this is all going to be equal to all going to be equal to 0 now that was interesting now we have an equation that has the derivative of Y with respect to x in it and this is what we essentially want to solve for this is the slope of the tangent line at any point so all we have to do at this point is solve for the derivative of Y with respect to X solve this equation so let's do that and actually just so we all so we can do this whole thing on the same page so we can see where we started let me copy and paste this up here this is where we left off and let's continue there so let's say let's subtract 2x from both sides so we're left with 2y times the derivative of Y with respect to X is equal to we're subtracting 2x from both sides so it's equal to negative 2x and then if we really want to solve for the derivative of Y with respect to X we can just divide both sides by 2i we just divide both sides by 2y and we are left with the derivative of Y with respect to X I'll scroll down a little bit the derivative of Y with respect to X is equal to is equal to well the twos cancel out we are left with negative x negative x over over y so this is interesting we didn't have to explicitly define y as a function of X here but we got our derivative in terms of an X and a y not just only in terms of an X but what does this mean well if we wanted to find let's say we wanted to find the derivative at this point this point right over here which if you're familiar with the unit circle so if this was a if this was a 45 degree angle this would be the square root of 2 over 2 comma the square root of 2 over 2 what is the slope of the tangent line there well we figured it out it's going to be the negative it's going to be negative x over Y so the slope of the tangent line here the slope of the tangent line right over here the slope is going to be equal to negative x so negative square root of 2 over 2 over Y over square root of 2 over 2 which is equal to negative 1 and it looks that looks just about right
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