Introduction to Implicit Differentiation

okay so today we are starting um this is like the last official technique of differentiation that we&#39;re going to be learning everything from this point on will either be an application or another area of calculus so this so this is it this is the home stretch this is not like it&#39;s relate or let&#39;s say use things that we&#39;ve learned about derivatives to do this but this process is very different than the way that we&#39;ve done it before so I&#39;m going to start off with just a quick little intro and let&#39;s see if you remember ready let&#39;s see if you remember if I write what is that what did I just put up there slope formula right that&#39;s a formula for slope um where that I could say that in words is in the numerator I have the difference in y and that is with respect to the denominator the difference in X right right when you subtract you get a difference um so if I kind of look at that the difference in y with respect to the difference in X what&#39;s that dydx is derivative right right and when we&#39;ve been finding derivatives it&#39;s allowed us to find the what of the tangent line the slope that&#39;s where I started sweet right right okay so we&#39;re going to be focusing on I&#39;m going to be using dydx notation today when we do these problems but this is just like kind of a reminder that this is what this is this is all connected this is all connected here so um specifically when we are taking these derivatives we are taking them with respect to X so let&#39;s just do a couple of um really basic examples to show you this process and then we&#39;ll jump into the procedure um so G make this a little bigger so for number one we have just a simple expression 3x2 if I want to take the derivative and that&#39;s what this notation means DDX means take the derivative of 3x^2 then what I am doing is the power rule which we know will give us 6X but I&#39;m taking the derivative of x with respect to X what&#39;s anything divided by itself one so DX DX cancels itself out and it just leaves me with 6X okay yeah are you still reporting the lessons I am okay let&#39;s try another one this one now has 2 y cubed so we&#39;re going to do the same thing take the derivative with respect to X of 2 y cubed again using the power rule I&#39;m going to get 6 y^2 but now I&#39;m taking the derivative of y with respect to X so this is not the same thing so I can&#39;t cancel it out it stays as part of uh the operation of taking the derivative okay um it&#39;s kind of like we use this dydx to sort of convert and that doesn&#39;t make full sense yet when we start to do some applications I&#39;ll point that out again um the fact that this is like a conversion and I&#39;ll I&#39;ll kind of show you how that works when we get a little further in let&#39;s try just one more simple one uh if I want to take the [Music] derivative of x plus the derivative with respect to X of 3 y I&#39;m going to get for the first term I&#39;m going to get one DX DX so we know that that cancels out and for the second term I&#39;m going to get three Dy DX so it just simplifies to 1 + 3 dydx yeah I&#39;m confused for number three you do the one and then DX over DX and then that cancels in plus how did you get dy over DX do you like multiply it or what because I&#39;m taking the derivative of a term that includes y with respect to X so it&#39;s the difference in y with respect to the difference X so the derivative is 3 Y which is three it&#39;s just 3 Dy DX just stay with me stay with me I know I know that this is this is new I know that this piece is new so just stay with me here okay okay one more and this one&#39;s going to involve product rule so remember our product rule um I am going to take and I&#39;m going to write this all out I&#39;m not going to do this for every problem that we do but I&#39;ll do it for this one so we take the first term times the derivative of the second term plus the second term times the derivative of uh the first factor I keep say term but I meant factors bless you so we&#39;re g to have x times the derivative of y^2 is 2 Y dydx and then plus the derivative of x is just 1 DX DX and again I&#39;m just going to cancel that out I&#39;m going to stop writing that very shortly so if I were to simplify this when I multiply X by 2 y right all of these things are multiplied together so I can put them in any order so I&#39;m going to write this as 2xy time dydx and then plus y^2 okay so the only new piece that we need to keep in mind as we do these problems is when we&#39;re taking the derivative of something that contains a y we have to include this this conversion factor dydx okay and that that&#39;ll become fully apparent uh as we do more of them and as we get into the applications you&#39;ll it&#39;s much easier to see when you have a problem with units why we do this okay so um up until this point all of the equations that we have taken derivatives of have been in the form y equals F ofx we might have used function notation we might have used Y and X but um either way we were able to solve for y that is called an um uh an equation that is defined explicitly because it&#39;s explicitly solved for y so the difference with what we&#39;re going to do today is the equations that we&#39;re going to work with um it&#39;s not impossible but it&#39;s not really easy to solve for y so I would not want to take this equation x^2 minus 2 y + 4 Y = 2 to try to solve that for y is a little bit messy and so So to avoid having to do that we um so that would be an equation that would be an equation that uh would be defined implicitly So to avoid having to solve an equation like this for y we&#39;re going to use this process of implicit differentiation um to find the derivative so this is going to look a lot different the the final answers that we get for these are in terms of X and Y so they look a lot different than the ones that we&#39;ve done before okay let&#39;s just do one example so I can show you the difference here so let&#39;s take the function um XY equals 1 I&#39;m going to do this one both ways to show you so the first way is explicitly and that means I&#39;m going to solve the equation for y this one is not too bad to solve for y again in some of them it&#39;s almost impossible but this one&#39;s doable so if I want to isolate y I&#39;m going to divide both sides by X and that is going to give me Y is equal to 1/x which we know from before if we wanted to take the derivative of that we would write that as X to the1 power right um if we apply the power rule to take the derivative then dydx is going to equal -1 time x^ -2 and then we would simplify this to be -1 over X squ oops so that would be our derivative explicitly that should not cause you any problem discomfort or anything that should be a no-brainer but now let&#39;s say that we did not want to solve that for why let&#39;s go through the process of implicitly finding the derivative so I&#39;m going to take the same equation same starting point x y = 1 and I want to take the derivative of the entire equation so I&#39;m going to do it both sides notice on the left hand side I have a product so I&#39;m going to write this one out again I&#39;m eventually going to stop writing this out you&#39;ll see but I&#39;m going to take the first Factor times the derivative of y plus y times the derivative of x and the derivative of one is what zero okay so I have that one simplified on the right hand side let&#39;s work on the left hand side the derivative of y it&#39;s going to be x * 1 Dy DX and then the derivative of x is just one DX DX I&#39;m not going to write the DX DX again so it&#39;s just y * 1 so this becomes just X Dy DX + y equals zero now I&#39;m trying to find dydx right so that&#39;s what I&#39;m trying to solve for so I&#39;m going to subtract a y from both sides and that&#39;s going to give me x dydx equal = y now I&#39;m going to divide both sides by an X and I will get dydx equals - y overx now you might say Mrs bian we didn&#39;t get the same answer when we did it explicitly that was our answer why isn&#39;t it the same over here somebody&#39;s got that question right shouldn&#39;t they be the same they actually are the same ready when we solved explicitly we said that y was equal to 1 overx so if I take this and substitute it in for y I will get 1 overx overx that&#39;s a complex fraction so it becomes ne1 /x times the reciprocal which is -1 over X2 it is the same as what we got over there it&#39;s just in a different form when you solve implicitly you don&#39;t have to change the form of your answer I only did that in this case to show you that you do get the same thing even though it might not look exactly the same okay all right everybody okay so far anybody okay so far okay all right let&#39;s go through some steps then first thing to remember which is what I kind of said at the very beginning if you are taking the derivative of any term that contains a y remember to multiply it by dydx so for now just remember to do this um yeah I said that a couple times so I&#39;m good here&#39;s the steps that you&#39;re going to do each time first thing we&#39;re always going to be differentiating um with respect to X someone in the other class asked do we ever differentiate with respect to Y generally when you have equations with X and Y right X is your independent variable Y is your dependent variable and so that&#39;s why we&#39;re differentiating with respect to X yes what does respect to X actually mean like that slope formula that we looked at at the beginning right you&#39;re finding the difference in y it&#39;s a ratio but that&#39;s in relation to or with respect to is another way to say that the denominator the change so the change that&#39;s happening in y occurs because of the change that is happening in X so it it&#39;s kind of like because of because of but they just say the mathematical way to say that is with respect to that&#39;s all that&#39;s all okay so so we don&#39;t flip-flop that we don&#39;t say the change in X with respect to the change in y because X is what sort of affects y um but we might use and we will use when we get to word problems um different variables so a lot of things will be with respect to time and that will be our independent variable so we might be differentiating with respect to T instead of with respect to X but it&#39;s it&#39;s always going to be your independent variable that is um what we use as like the basis okay um so next step once we do that part of it we get all of the term oops we get all of the terms that have a dydx onto one side or the other of the equation just put them all together we&#39;re going to factor out a dydx and then we&#39;re going to solve for dydx so it&#39;s those same steps every time we do this it&#39;s just a matter of the algebra that is involved with doing it and honestly the algebra is what trips people up with this kind of like the EXA we won&#39;t go there okay so let&#39;s do um this one this example is actually easier than the first one that we did okay because the first one that we did had the product rule in it this one is just the summon difference of a bunch of terms so we&#39;re going to differentiate each term with respect to X so if I take this whole equation and take the derivative with respect to X I&#39;m going to get 3 y^ 2 Dy DX Plus 2y Dy DX minus 5 Dy DX minus 2x DX DX which I&#39;m going to stop writing now I&#39;m not writing that anymore because that just disappears right and then on the other side of the equal sign sign the derivative of a constant is zero yeah why do we have to write the Dy DX because we&#39;re differentiating each term with respect to X so that so for each term we&#39;re including this and we didn&#39;t have to do this when it was solved explicitly because when you took when you had an equation let&#39;s just say yal 2X and we wanted to take the derivative of that the derivative of y Y is dydx that&#39;s what we were using as our notation and then on the other side we were taking the derivative and it was two DX DX now we didn&#39;t write all of that out last time but this was an unnecessary step when the equation was solved explicitly it&#39;s showing up now because of our implicit definition okay okay now this is unusual that all of the Dy DXs are on the left hand side so this is really really easy so I&#39;m just going to I&#39;m going to do two things at the same time I don&#39;t think this is going to confuse you but I&#39;m first going to factor out a dydx and I&#39;m going to be left with 3 y^ 2 + 2 y - five and then at the same time I&#39;m going to add my two X over to the other side so that I just get uh 2x there okay do everybody say I did two things at the same time that these problems involve a lot of writing so I just try to save so I don&#39;t have to write every step like 8,000 times um so you can combine things just be careful when you do it just be careful I&#39;m telling you it&#39;s the algebra that gets people on this not anything else so now dydx and then those parentheses mean multiplication how do you undo multiplication division so I&#39;m G now I&#39;m not going to write this because again it&#39;s more writing but I&#39;m going to divide now on both sides and that&#39;s going to isolate my dydx which is what I&#39;m trying to do I have the 2X and then it&#39;s going to be over 3 y^ 2 + 2 y minus 5 so notice a couple of things first of all um notice my use of parentheses here when I factor out um parentheses are going to be your friend in these problems if you don&#39;t use them you are going to confuse multip ication with addition and subtraction so I would error on the side of using them also notice what my answer looks like this does not look like any derivative that we&#39;ve calculated up until this point again my answer is going to be in terms of X and Y when I isolate this dydx okay so that is the first basic type of problem with this any questions on that one before before I move on yes are you so are you always going to factor out a Dy always it&#39;s like let&#39;s say it&#39;s like 3 y^ 2+ 9 Yus 3 instead of those then would youor well then you&#39;re not Factor then you don&#39;t need to factor because there&#39;s only one term that has it but that&#39;s going to be the exception that that&#39;s only going to happen occasionally most cases you&#39;re going to have at least two terms if not more have a dydx in okay so if you only have one then you&#39;re just going to isolate by dividing by whatever the value is in it okay let&#39;s try one uh harder one that involves the product rule so number six we are going to do the same start the same way oops we are going to take the derivative of the whole equation with respect to X and let&#39;s just do the left side first so this becomes 3x 2ar DX DX minus 3 y^ 2 Dy DX and now we deal with our product so this 6X will call our first factor and the Y will be our second the derivative of 6X is 6 DX DX the derivative of y I&#39;m gonna just move this down a little bit the derivative of y is one Dy DX and I&#39;m I just wrote that there so when I do the product rule it&#39;s a little bit easier for me to see so I&#39;m going to take first Factor 6X times the derivative of the second Factor one dydx is just dydx I don&#39;t have to write the one plus the Y times the derivative of 6X which is just 6 so now notice that I have a dydx on each side of the equation so I&#39;m going to have to move things around here let me simplify this first and just get this to be 3x^2 - 3 y^2 dydx = 6X dydx + 6 y so now I&#39;m to do but you where I got with this 6X right here yeah because when I do product so I&#39;m counting this as the product of 6xtimes Y so that was my first ter or first Factor okay I&#39;m going to do two steps again at the same time here so um I don&#39;t like to have negatives in the front so I&#39;m going to add this one over to this side and then simultaneously I&#39;m going to subtract this and bring it over to this side so I&#39;m going to show you can do this in two steps if you want but I&#39;m going to do minus 6 Y and I&#39;m going to do plus 3 y^2 dydx okay I&#39;m not trying to confuse you I&#39;m just again trying to save writing so that is going to now give me on the left 3x^2 uh - 6 y equals um both terms on this side are going to have a dydx right because this one&#39;s already here and I&#39;m moving this one over they both have dydx I&#39;m gonna pull that out factor that out so it becomes dydx times I&#39;m going to have a positive 3 y^ 2 plus 6X again you could write that as two different steps the factoring out if you wanted to everybody with me though with what I did hopefully this again means multiplication so I&#39;m going to divide both sides by this expression and I&#39;m going to end up with I&#39;m just going to reverse the sides so dydx is equal to my numerator that I already had over what I&#39;m dividing by on both sides this is what I&#39;m dividing by often but not always this can further be further simplified so notice how do I say this I would highly encourage you so that you don&#39;t just start cancelling things right this is subtraction and addition we don&#39;t want to be killing anything I would highly recommend that you factor out before you try to cancel so that you don&#39;t cancel terms and that you&#39;re cancelling factors so my advice to you is to take the step because these get complicated force yourself to bring out the GCF in both the numerator and the denominator and then you can see that your threes are going to cancel so you&#39;re going to end up with dydx is x² - 2 y over y^ 2 + 2x okay little off okay [Music] questions [Music] um when do we end how did I get okay let&#39;s back up everybody good let me at least do what as far as I can get um notice the first problem first example that we did was just some indifference right there was that was pretty basic each we just just took the derivative of each term notice the second problem involved a product so we use product rule when we did the derivative on the right hand side when you look at number seven what rule for derivatives would you think of quotient rule quotient rule okay let&#39;s start let&#39;s see let&#39;s take the derivative of both sides with respect to X now before I apply the quotient rule I&#39;m going to write the derivative of my numerator is now be careful here the derivative of my numerator is 2 DX DX + 1 dydx and the derivative of my denominator is 1 DX DX - 5 dydx now I just wrote that in there so that when I do the quotient rule I have the pieces that I need to be able to do it okay so here we go quotient rule low wrap everything in parentheses dhide minus Pi oops DLo over um x - 5 y quantity squared yikes anybody want to do all that foiling I don&#39;t I don&#39;t want to do all that foiling anybody want to see a better way to maybe do this let&#39;s take the original problem and let&#39;s I can write one as a fraction one over one right well now that&#39;s a proportion we can solve proportions by cross multiplying can&#39;t we so we get my pen is done 2x + y = x - 5 y well now that&#39;s super easy now you just take the derivative of each term and you follow the process that we did um with that first problem so if I had to do this problem I would definitely not use quotient problem you certainly can finish that and you will get the same answer but it&#39;s going to be a lot Messier lot more chance for errors I would approach it this way I&#39;m not gonna have time to finish this example so I&#39;ll pick this up tomorrow we&#39;ll finish this one in class tomorrow before you go I&#39;m going to give you your homework

okay so today we are starting um this is like the last official technique of differentiation that we&amp;#39;re going to be learning everything from this point on will either be an application or another area of calculus so this so this is it this is the home stretch this is not like it&amp;#39;s relate or let&amp;#39;s say use things that we&amp;#39;ve learned about derivatives to do this but this process is very different than the way that we&amp;#39;ve done it before so I&amp;#39;m going to start off with just a quick little intro and let&amp;#39;s see if you remember ready let&amp;#39;s see if you remember if I write what is that what did I just put up there slope formula right that&amp;#39;s a formula for slope um where that I could say that in words is in the numerator I have the difference in y and that is with respect to the denominator the difference in X right right when you subtract you get a difference um so if I kind of look at that the difference in y with respect to the difference in X what&amp;#39;s that dydx is derivative right right and when we&amp;#39;ve been finding derivatives it&amp;#39;s allowed us to find the what of the tangent line the slope that&amp;#39;s where I started sweet right right okay so we&amp;#39;re going to be focusing on I&amp;#39;m going to be using dydx notation today when we do these problems but this is just like kind of a reminder that this is what this is this is all connected this is all connected here so um specifically when we are taking these derivatives we are taking them with respect to X so let&amp;#39;s just do a couple of um really basic examples to show you this process and then we&amp;#39;ll jump into the procedure um so G make this a little bigger so for number one we have just a simple expression 3x2 if I want to take the derivative and that&amp;#39;s what this notation means DDX means take the derivative of 3x^2 then what I am doing is the power rule which we know will give us 6X but I&amp;#39;m taking the derivative of x with respect to X what&amp;#39;s anything divided by itself one so DX DX cancels itself out and it just leaves me with 6X okay yeah are you still reporting the lessons I am okay let&amp;#39;s try another one this one now has 2 y cubed so we&amp;#39;re going to do the same thing take the derivative with respect to X of 2 y cubed again using the power rule I&amp;#39;m going to get 6 y^2 but now I&amp;#39;m taking the derivative of y with respect to X so this is not the same thing so I can&amp;#39;t cancel it out it stays as part of uh the operation of taking the derivative okay um it&amp;#39;s kind of like we use this dydx to sort of convert and that doesn&amp;#39;t make full sense yet when we start to do some applications I&amp;#39;ll point that out again um the fact that this is like a conversion and I&amp;#39;ll I&amp;#39;ll kind of show you how that works when we get a little further in let&amp;#39;s try just one more simple one uh if I want to take the [Music] derivative of x plus the derivative with respect to X of 3 y I&amp;#39;m going to get for the first term I&amp;#39;m going to get one DX DX so we know that that cancels out and for the second term I&amp;#39;m going to get three Dy DX so it just simplifies to 1 + 3 dydx yeah I&amp;#39;m confused for number three you do the one and then DX over DX and then that cancels in plus how did you get dy over DX do you like multiply it or what because I&amp;#39;m taking the derivative of a term that includes y with respect to X so it&amp;#39;s the difference in y with respect to the difference X so the derivative is 3 Y which is three it&amp;#39;s just 3 Dy DX just stay with me stay with me I know I know that this is this is new I know that this piece is new so just stay with me here okay okay one more and this one&amp;#39;s going to involve product rule so remember our product rule um I am going to take and I&amp;#39;m going to write this all out I&amp;#39;m not going to do this for every problem that we do but I&amp;#39;ll do it for this one so we take the first term times the derivative of the second term plus the second term times the derivative of uh the first factor I keep say term but I meant factors bless you so we&amp;#39;re g to have x times the derivative of y^2 is 2 Y dydx and then plus the derivative of x is just 1 DX DX and again I&amp;#39;m just going to cancel that out I&amp;#39;m going to stop writing that very shortly so if I were to simplify this when I multiply X by 2 y right all of these things are multiplied together so I can put them in any order so I&amp;#39;m going to write this as 2xy time dydx and then plus y^2 okay so the only new piece that we need to keep in mind as we do these problems is when we&amp;#39;re taking the derivative of something that contains a y we have to include this this conversion factor dydx okay and that that&amp;#39;ll become fully apparent uh as we do more of them and as we get into the applications you&amp;#39;ll it&amp;#39;s much easier to see when you have a problem with units why we do this okay so um up until this point all of the equations that we have taken derivatives of have been in the form y equals F ofx we might have used function notation we might have used Y and X but um either way we were able to solve for y that is called an um uh an equation that is defined explicitly because it&amp;#39;s explicitly solved for y so the difference with what we&amp;#39;re going to do today is the equations that we&amp;#39;re going to work with um it&amp;#39;s not impossible but it&amp;#39;s not really easy to solve for y so I would not want to take this equation x^2 minus 2 y + 4 Y = 2 to try to solve that for y is a little bit messy and so So to avoid having to do that we um so that would be an equation that would be an equation that uh would be defined implicitly So to avoid having to solve an equation like this for y we&amp;#39;re going to use this process of implicit differentiation um to find the derivative so this is going to look a lot different the the final answers that we get for these are in terms of X and Y so they look a lot different than the ones that we&amp;#39;ve done before okay let&amp;#39;s just do one example so I can show you the difference here so let&amp;#39;s take the function um XY equals 1 I&amp;#39;m going to do this one both ways to show you so the first way is explicitly and that means I&amp;#39;m going to solve the equation for y this one is not too bad to solve for y again in some of them it&amp;#39;s almost impossible but this one&amp;#39;s doable so if I want to isolate y I&amp;#39;m going to divide both sides by X and that is going to give me Y is equal to 1/x which we know from before if we wanted to take the derivative of that we would write that as X to the1 power right um if we apply the power rule to take the derivative then dydx is going to equal -1 time x^ -2 and then we would simplify this to be -1 over X squ oops so that would be our derivative explicitly that should not cause you any problem discomfort or anything that should be a no-brainer but now let&amp;#39;s say that we did not want to solve that for why let&amp;#39;s go through the process of implicitly finding the derivative so I&amp;#39;m going to take the same equation same starting point x y = 1 and I want to take the derivative of the entire equation so I&amp;#39;m going to do it both sides notice on the left hand side I have a product so I&amp;#39;m going to write this one out again I&amp;#39;m eventually going to stop writing this out you&amp;#39;ll see but I&amp;#39;m going to take the first Factor times the derivative of y plus y times the derivative of x and the derivative of one is what zero okay so I have that one simplified on the right hand side let&amp;#39;s work on the left hand side the derivative of y it&amp;#39;s going to be x * 1 Dy DX and then the derivative of x is just one DX DX I&amp;#39;m not going to write the DX DX again so it&amp;#39;s just y * 1 so this becomes just X Dy DX + y equals zero now I&amp;#39;m trying to find dydx right so that&amp;#39;s what I&amp;#39;m trying to solve for so I&amp;#39;m going to subtract a y from both sides and that&amp;#39;s going to give me x dydx equal = y now I&amp;#39;m going to divide both sides by an X and I will get dydx equals - y overx now you might say Mrs bian we didn&amp;#39;t get the same answer when we did it explicitly that was our answer why isn&amp;#39;t it the same over here somebody&amp;#39;s got that question right shouldn&amp;#39;t they be the same they actually are the same ready when we solved explicitly we said that y was equal to 1 overx so if I take this and substitute it in for y I will get 1 overx overx that&amp;#39;s a complex fraction so it becomes ne1 /x times the reciprocal which is -1 over X2 it is the same as what we got over there it&amp;#39;s just in a different form when you solve implicitly you don&amp;#39;t have to change the form of your answer I only did that in this case to show you that you do get the same thing even though it might not look exactly the same okay all right everybody okay so far anybody okay so far okay all right let&amp;#39;s go through some steps then first thing to remember which is what I kind of said at the very beginning if you are taking the derivative of any term that contains a y remember to multiply it by dydx so for now just remember to do this um yeah I said that a couple times so I&amp;#39;m good here&amp;#39;s the steps that you&amp;#39;re going to do each time first thing we&amp;#39;re always going to be differentiating um with respect to X someone in the other class asked do we ever differentiate with respect to Y generally when you have equations with X and Y right X is your independent variable Y is your dependent variable and so that&amp;#39;s why we&amp;#39;re differentiating with respect to X yes what does respect to X actually mean like that slope formula that we looked at at the beginning right you&amp;#39;re finding the difference in y it&amp;#39;s a ratio but that&amp;#39;s in relation to or with respect to is another way to say that the denominator the change so the change that&amp;#39;s happening in y occurs because of the change that is happening in X so it it&amp;#39;s kind of like because of because of but they just say the mathematical way to say that is with respect to that&amp;#39;s all that&amp;#39;s all okay so so we don&amp;#39;t flip-flop that we don&amp;#39;t say the change in X with respect to the change in y because X is what sort of affects y um but we might use and we will use when we get to word problems um different variables so a lot of things will be with respect to time and that will be our independent variable so we might be differentiating with respect to T instead of with respect to X but it&amp;#39;s it&amp;#39;s always going to be your independent variable that is um what we use as like the basis okay um so next step once we do that part of it we get all of the term oops we get all of the terms that have a dydx onto one side or the other of the equation just put them all together we&amp;#39;re going to factor out a dydx and then we&amp;#39;re going to solve for dydx so it&amp;#39;s those same steps every time we do this it&amp;#39;s just a matter of the algebra that is involved with doing it and honestly the algebra is what trips people up with this kind of like the EXA we won&amp;#39;t go there okay so let&amp;#39;s do um this one this example is actually easier than the first one that we did okay because the first one that we did had the product rule in it this one is just the summon difference of a bunch of terms so we&amp;#39;re going to differentiate each term with respect to X so if I take this whole equation and take the derivative with respect to X I&amp;#39;m going to get 3 y^ 2 Dy DX Plus 2y Dy DX minus 5 Dy DX minus 2x DX DX which I&amp;#39;m going to stop writing now I&amp;#39;m not writing that anymore because that just disappears right and then on the other side of the equal sign sign the derivative of a constant is zero yeah why do we have to write the Dy DX because we&amp;#39;re differentiating each term with respect to X so that so for each term we&amp;#39;re including this and we didn&amp;#39;t have to do this when it was solved explicitly because when you took when you had an equation let&amp;#39;s just say yal 2X and we wanted to take the derivative of that the derivative of y Y is dydx that&amp;#39;s what we were using as our notation and then on the other side we were taking the derivative and it was two DX DX now we didn&amp;#39;t write all of that out last time but this was an unnecessary step when the equation was solved explicitly it&amp;#39;s showing up now because of our implicit definition okay okay now this is unusual that all of the Dy DXs are on the left hand side so this is really really easy so I&amp;#39;m just going to I&amp;#39;m going to do two things at the same time I don&amp;#39;t think this is going to confuse you but I&amp;#39;m first going to factor out a dydx and I&amp;#39;m going to be left with 3 y^ 2 + 2 y - five and then at the same time I&amp;#39;m going to add my two X over to the other side so that I just get uh 2x there okay do everybody say I did two things at the same time that these problems involve a lot of writing so I just try to save so I don&amp;#39;t have to write every step like 8,000 times um so you can combine things just be careful when you do it just be careful I&amp;#39;m telling you it&amp;#39;s the algebra that gets people on this not anything else so now dydx and then those parentheses mean multiplication how do you undo multiplication division so I&amp;#39;m G now I&amp;#39;m not going to write this because again it&amp;#39;s more writing but I&amp;#39;m going to divide now on both sides and that&amp;#39;s going to isolate my dydx which is what I&amp;#39;m trying to do I have the 2X and then it&amp;#39;s going to be over 3 y^ 2 + 2 y minus 5 so notice a couple of things first of all um notice my use of parentheses here when I factor out um parentheses are going to be your friend in these problems if you don&amp;#39;t use them you are going to confuse multip ication with addition and subtraction so I would error on the side of using them also notice what my answer looks like this does not look like any derivative that we&amp;#39;ve calculated up until this point again my answer is going to be in terms of X and Y when I isolate this dydx okay so that is the first basic type of problem with this any questions on that one before before I move on yes are you so are you always going to factor out a Dy always it&amp;#39;s like let&amp;#39;s say it&amp;#39;s like 3 y^ 2+ 9 Yus 3 instead of those then would youor well then you&amp;#39;re not Factor then you don&amp;#39;t need to factor because there&amp;#39;s only one term that has it but that&amp;#39;s going to be the exception that that&amp;#39;s only going to happen occasionally most cases you&amp;#39;re going to have at least two terms if not more have a dydx in okay so if you only have one then you&amp;#39;re just going to isolate by dividing by whatever the value is in it okay let&amp;#39;s try one uh harder one that involves the product rule so number six we are going to do the same start the same way oops we are going to take the derivative of the whole equation with respect to X and let&amp;#39;s just do the left side first so this becomes 3x 2ar DX DX minus 3 y^ 2 Dy DX and now we deal with our product so this 6X will call our first factor and the Y will be our second the derivative of 6X is 6 DX DX the derivative of y I&amp;#39;m gonna just move this down a little bit the derivative of y is one Dy DX and I&amp;#39;m I just wrote that there so when I do the product rule it&amp;#39;s a little bit easier for me to see so I&amp;#39;m going to take first Factor 6X times the derivative of the second Factor one dydx is just dydx I don&amp;#39;t have to write the one plus the Y times the derivative of 6X which is just 6 so now notice that I have a dydx on each side of the equation so I&amp;#39;m going to have to move things around here let me simplify this first and just get this to be 3x^2 - 3 y^2 dydx = 6X dydx + 6 y so now I&amp;#39;m to do but you where I got with this 6X right here yeah because when I do product so I&amp;#39;m counting this as the product of 6xtimes Y so that was my first ter or first Factor okay I&amp;#39;m going to do two steps again at the same time here so um I don&amp;#39;t like to have negatives in the front so I&amp;#39;m going to add this one over to this side and then simultaneously I&amp;#39;m going to subtract this and bring it over to this side so I&amp;#39;m going to show you can do this in two steps if you want but I&amp;#39;m going to do minus 6 Y and I&amp;#39;m going to do plus 3 y^2 dydx okay I&amp;#39;m not trying to confuse you I&amp;#39;m just again trying to save writing so that is going to now give me on the left 3x^2 uh - 6 y equals um both terms on this side are going to have a dydx right because this one&amp;#39;s already here and I&amp;#39;m moving this one over they both have dydx I&amp;#39;m gonna pull that out factor that out so it becomes dydx times I&amp;#39;m going to have a positive 3 y^ 2 plus 6X again you could write that as two different steps the factoring out if you wanted to everybody with me though with what I did hopefully this again means multiplication so I&amp;#39;m going to divide both sides by this expression and I&amp;#39;m going to end up with I&amp;#39;m just going to reverse the sides so dydx is equal to my numerator that I already had over what I&amp;#39;m dividing by on both sides this is what I&amp;#39;m dividing by often but not always this can further be further simplified so notice how do I say this I would highly encourage you so that you don&amp;#39;t just start cancelling things right this is subtraction and addition we don&amp;#39;t want to be killing anything I would highly recommend that you factor out before you try to cancel so that you don&amp;#39;t cancel terms and that you&amp;#39;re cancelling factors so my advice to you is to take the step because these get complicated force yourself to bring out the GCF in both the numerator and the denominator and then you can see that your threes are going to cancel so you&amp;#39;re going to end up with dydx is x² - 2 y over y^ 2 + 2x okay little off okay [Music] questions [Music] um when do we end how did I get okay let&amp;#39;s back up everybody good let me at least do what as far as I can get um notice the first problem first example that we did was just some indifference right there was that was pretty basic each we just just took the derivative of each term notice the second problem involved a product so we use product rule when we did the derivative on the right hand side when you look at number seven what rule for derivatives would you think of quotient rule quotient rule okay let&amp;#39;s start let&amp;#39;s see let&amp;#39;s take the derivative of both sides with respect to X now before I apply the quotient rule I&amp;#39;m going to write the derivative of my numerator is now be careful here the derivative of my numerator is 2 DX DX + 1 dydx and the derivative of my denominator is 1 DX DX - 5 dydx now I just wrote that in there so that when I do the quotient rule I have the pieces that I need to be able to do it okay so here we go quotient rule low wrap everything in parentheses dhide minus Pi oops DLo over um x - 5 y quantity squared yikes anybody want to do all that foiling I don&amp;#39;t I don&amp;#39;t want to do all that foiling anybody want to see a better way to maybe do this let&amp;#39;s take the original problem and let&amp;#39;s I can write one as a fraction one over one right well now that&amp;#39;s a proportion we can solve proportions by cross multiplying can&amp;#39;t we so we get my pen is done 2x + y = x - 5 y well now that&amp;#39;s super easy now you just take the derivative of each term and you follow the process that we did um with that first problem so if I had to do this problem I would definitely not use quotient problem you certainly can finish that and you will get the same answer but it&amp;#39;s going to be a lot Messier lot more chance for errors I would approach it this way I&amp;#39;m not gonna have time to finish this example so I&amp;#39;ll pick this up tomorrow we&amp;#39;ll finish this one in class tomorrow before you go I&amp;#39;m going to give you your homework

Transcript for:Introduction to Implicit Differentiation

Transcript for:
Introduction to Implicit Differentiation