in section 3.8 we're gonna cover the topic of implicit differentiation so to this point we have been working with functions strictly in what we call explicit form so for example here I have this quadratic polynomial so we might say y is equal to x squared plus 2x plus 1 so we call this explicit form because the connection between x and y is very clear very explicit you might say so for instance in order to find my Y value I have the recipe and the recipe is completely in terms of X's or constants which do not involve a variable so for instance to find my Y value my Y value is simply x squared plus 2x plus 1 so if X is equal to 1 we'll find a corresponding Y value of 4 so this is what we'll call a function in explicit form it's y solved completely in terms of X now we know how to find the derivative now the derivative of this expression by the power rule is very simply 2x plus 2 now the question is how do we find the derivative if we do not have implicit form available excuse me explicit form available to us in other words we have what we call implicit form so notice here that in this equation the X's and the Y's are sort of mixed together and I do not have an expression like this where I can say Y is equal to some expression involving my X terms so I have x squared plus y squared equals x times y this is either very difficult or impossible to solve for y and so I do not want to have to rely on this explicit form in order to find a derivative I would like to be able to find the derivative dy/dx even if I have this implicit form given to me so the two differences here again in explicit form we have y equals use some expression involving our X terms now the X's and the Y's are convoluted they're mixed together and I don't have an expression read available that I can turn into an you know why equals type of equation and then find the derivative so implicit differentiation is going to tell us how we find the derivative dy/dx when we have this implicit form given to us so if Y is not if Y is a function of X how will we find dy/dx if the explicit form is not given or not possible to State and that's the technique of implicit differentiation so let's look at this equation x squared plus y squared equals 1 now this is just a circle of radius 1 centered at the origin so the key step here is that we're going to differentiate both sides of the equation and we have to remember to apply the chain rule every time we take the derivative of a Y term since Y is a function of X Y has a dependency on X and that's exactly the case when we have to use the chain rule when we have a function that has a dependency on some other some other object so when I take the derivative of a Y term that derivative of the Y term is going to pick up a dy/dx so notice here we'll take the derivative of both sides of the equation the derivative of this left hand side x squared plus y squared well X terms are going to behave just as we think they should so the derivative of x squared is just going to be 2x the derivative of Y squared now we're going to use the power rule so we'll bring down the 2 take one away so I have a 2y but that Y is a function of X I have to tack on dy/dx and the derivative of the right hand side is equal to zero instead of dy/dx you're free to use Y prime if you prefer now the rest of this is just an algebra problem I'd like to solve for dy/dx so solving for dy/dx I'm going to move this to x over and then divide by 2 y and so my derivative dy/dx we will express as negative x over Y and this is the technique of implicit differentiation so for a simple example look at this equation here y minus x squared plus 2x equals 1 this is very easily solved for y and then we find the derivative to be 2x minus 2 well let's say we don't solve for y we can apply this technique of implicit differentiation the derivative of Y is going to be 1 but Y is the function of X so I have 1 times dy/dx and now the derivative of minus x squared is minus 2x derivative of 2x is 2 derivative of 1 is 0 rearranging some things we see that we have exactly the same expression 2x minus 2 so there's nothing special about this technique the method is working just as we expect it would for functions that we can solve for y this is how this method is going to be critical when we have expressions that cannot be solved for y and I want to find the derivative so let's look at this equation y squared plus 3x equals 2 and we'll find the derivative at this point negative 1 root 5 now these problems will sometimes give you a point instead of just an x value because the derivative will often involve not always but the derivative will often involve an x and a y so there's multiple things that have to be plugged into the derivative my equation is y squared plus 3x equals 2 now notice that this can be solved for y but because of the square I'm gonna pick up this plus or minus and my derivative has this might if I take the derivative through this through this direction I've got this plus or minus that has to hang out with the derivative we can avoid that by just applying the technique of implicit differentiation so we're gonna just apply the technique here jumping right in the derivative of Y squared is going to be 2 Y but Y is a function of X so I get 2y dy/dx the derivative of 3x is 3 and the derivative of 2 is 0 so that's the whole calculus right here the rest of this is just solving for dy/dx so what we're gonna have is 2y dy/dx we'll subtract the 3 over so we'll get negative 3 and then dividing by 2y here's the expression for the derivative negative 3 divided by 2y and now we're going to find the derivative or the slope of the tangent line at this point 1 negative 1 root 5 now in this case notice the derivative has no X involved so the derivative does not depend on X here and so we'll just plug in root 5 so at this point negative 1 root 5 the slope of the tangent line will be negative 3 over 2 root 5 so let's look at this function or this equation YX squared plus y cubed equals cosine Y and I want to find dy/dx now again notice we're not solved for y and solving for y less like something that's going to be a difficult proposition so no problem we're just going to use implicit differentiation to find dy/dx now notice on this left hand side this right here is a product so we have to still apply the rules that we know I have to apply the product rule on that term so on the left hand side we'll jump right into the derivative the derivative of YX squared will apply the product rule derivative of Y is going to be dy DX x squared and then plus y times the derivative of x squared which is 2x now the derivative of Y cubed is 3 y squared dy DX and on the right hand side the derivative of cosine is negative sine but since we have a Y involved we're gonna pick up dy/dx so that's all of the calculus here again the rest of this is just solving isolating dy/dx so let's move some things around so we'll have dy/dx x squared plus 3 y squared dy DX we'll move that negative sign term over we'll get positive sine of Y dy DX and we'll take this Y times 2x move it to the right hand side I'm gonna have negative 2xy so let's just factor the dy/dx out now dy/dx now we have x squared plus 3 y squared plus the sine of Y and that's going to equal negative 2xy and to isolate D we will divide by everything we see in parentheses so I have negative 2x Y over x squared plus 3y squared plus the sine of Y and that expression there as bad as it looks is our expression for the derivative dy/dx for another example here's x squared y squared plus e to the y equals the tangent of Y so again notice we're not solved for y and it looks like it would be very difficult to even do so if not impossible so we're gonna jump right in to implicit differentiation notice again I have a product here so applying the product rule my derivative will be 2xy squared plus x squared the derivative of Y squared will be 2y but since it's a function of X I tack on dy/dx now the derivative of e to the Y is e to the Y but again it involves the Y so I tack on dy/dx now on the right hand side derivative of tangent is secant squared of Y but again we have a Y involved attack on a dy/dx so let's now just solve for dy/dx so what we'll do here let me write the dy/dx is all on the left hand side now this is your choice here you could write the dy/dx is all on the right hand side so I have x squared Y times dy/dx plus e to the Y dy DX we'll move that secant squared of Y over and on the right hand side I'll have negative 2xy squared taking this term and moving it over let's factor the dy/dx is out so I'll get x squared y plus e to the Y minus secant squared equals negative 2 XY squared and then we'll isolate dy DX divided by the paren thing in parentheses so I have negative 2 XY squared over x squared y plus e to the Y minus secant squared of Y and that expression dydx is my derivative so let's use this idea now to find the equation of a normal line so at the point 1 1 I'd like the normal line for this curve or this implicitly defined function X cubed plus x squared y plus 4 y squared equals 6 so we'll jump right in again we're not solved for y so this suggests implicit differentiation the derivative of x cubed is 3x squared now that middle term x squared Y is a product so I'll have 2 X Y and then plus x squared times the derivative of Y which is 1 but Y is a function of X so it's 1 dy DX now for y squared the derivative will be 8 Y dy DX derivative of the right hand side 6 is just 0 now I'm just looking for the derivative or this the equation of the normal line at the point 1 1 so I'm going to take this expression here and just plug in x equals 1 and y equals 1 and solve for dy DX I'm not really interested in the derivative itself just at a particular point so plugging in x equals 1 y equals 1 we're going to have 3 plus 2 plus dy/dx plus 8 dy/dx now notice this is still equal to 0 notice 3 & 2 is 5 so I have 5 and now I'm going to have 9 dy DX's and I can solve for dy/dx subtract 5 divided by 9 so my derivative at the point 1 1 is negative 5 over 9 so that tells me the slope of the normal line is positive 9 over 5 and now we can put the line together we'll have Y minus our y-coordinate of 1 equals our slope of 9 fifths times X minus the x-coordinate and we will leave this in point-slope form how about this implicit equation here one plus x equals the sine of X Y squared so let's find the derivative so on the left hand side the derivative is very simply going to be 1 now on the right hand side the derivative of sine is cosine but I have cosine of X Y squared and now I have to take the derivative of the inside now notice that's a product so the derivative of XY squared using the product rule I'll have Y squared plus X times 2y dy/dx so there's the expression for the derivative the first step at the expression for the derivative dy/dx now I'd like to isolate what I'll do here there's a couple of ways of approaching this but we could distribute this term here and then start to simplify by isolating dy/dx I'm going to go ahead and just divide by cosine XY squared off the top so I'll have 1 over cosine XY squared and that's going to equal Y squared I'm gonna write this as 2 XY dy/dx now on the right hand side on the left hand side 1 over cosine that's the same as secant so this is secant XY squared I'm gonna move that Y squared over and then have 2 XY dy/dx and now I can isolate dy/dx just by dividing by 2 XY so our derivative is going to be secant XY squared minus y squared divide it by 2 XY and that expression is our derivative dy/dx as one last example let's show that these two curves intersect at a right angle so I have a hyperbola the hyperbola is defined this way x squared minus y squared equals 5 and I have an ellipse F 4x squared plus 9y squared equals 72 these two intersect at 3/2 and I want to show that that intersection is happening at a right angle so the way we'll do this is that we will show the tangent lines at that point are perpendicular if lines are perpendicular their slopes are negative reciprocals so that's all we want to do here show that the two derivatives are negative reciprocals of each other at that point 3/2 so for the first part for the hyperbola our derivative will be 2x minus 2y dy/dx equals 0 and we're gonna plug in 2 0.32 so x equals 3 y equals 2 we'll have 6 minus 4 dy/dx equals 0 so 6 is going to equal 4 dy/dx dividing by 4 we'll get dy/dx is 6 over 4 which is 3 over 2 so for the ellipse I'd like to see negative 2/3 that will show me that they're intersecting at a right angle so implicit differentiation I'll have 8x plus 18 y dy DX equals 0 again plugging in x equals 3 and y equals 2 this is the equation I get so dy DX well 36 dy DX is going to be equal to negative 24 all right so we have dy DX is going to be negative 24 over 36 dividing by 12 we get negative 2/3 which is exactly the negative reciprocal of 3 halves so we are seeing that these two curves intersect at a right angle because the tangent lines are perpendicular at those two at that particular point 3 2 one last note before we leave this section sometimes we have to be careful with implicit equations because not every implicit equation will be meaningful so here's an example of what we mean by that so why might the derivative dy/dx for this equation e to the x squared plus y squared equals zero be quote-unquote junk so the idea here is that every implicit equation we have an underlying assumption that there's some function that exists which satisfies the equation but you should notice here that e to some power is never going to be equal to 0 there is no combination of x and y here that will ever give me e to this expression being equal to 0 so in this case there is no function that satisfies this particular equation so although we could go through the steps to find the derivative through the technique of implicit differentiation that derivative is going to represent a function which which fails to be defined or fails to exist because e to some power will never be equal to 0 so in this case there's no function that satisfies this equation so the derivative represents a function which is undefined or a function we might say that just doesn't exist so not every implicit equation is meaningful in the fact that not every implicit equation has a function which will satisfy it and so we want to be careful we don't really have to think about this a lot but it's something you should be aware of that there is this assumption baked into these particular types of equations so so just a final few notes here for a summary so I will let you read through these and this will in the lecture for implicit differentiation