 differentiation of implicit functions or functions which are defined implicitly so first of all what is the meaning of an implicit function so implicit function is basically where your variables x and y cannot be separated out easily see up till now what happened most of the functions that you got where of the type y is equal to f of x where there was a clear cut or you could easily separate the two variables from each other so the independent variable x and the dependent variable y terms could be separated out from each other so finding dy by dx was easy so this is called explicitly defined function this is called explicitly defined function explicitly defined function so today we are going to talk about implicitly defined function where your x and y separation is very very difficult so I cannot write y as a function of x so it's basically a mixed function of x and y okay equal to zero or equal to constant whatever you want to say okay so this is a implicitly defined function implicitly defined function implicitly defined function to give you some examples for the same let's say I have a function like this I have a function like this sine square y plus cos of x y equal to 1 so this is an example of an implicitly defined function so how do we find dy by dx in this case that is what we are going to learn how to find dy by dx in this case that's what we are going to learn in this particular expression so what I'm going to do here is I'm going to let's say call this term let's say I have this function and I want to find dy by dx okay so we'll call this term as let's say u okay let's say I call this term as a u okay now let's say I differentiate both sides with respect to x I differentiate both sides with respect to x okay now see how this works how this works if you differentiate this with respect to t let's not write x as of now let me just give you a very generic idea if you differentiate this with respect to t du by dt will be please make a note of this this is something which is beyond your school level syllabus it is doh f by doh x into dx by dt plus doh f by doh y into dy by dt basically what you have used here is actually a complete form of a chain rule this is called total differential so if you're differentiating you with respect to t and since this function since this function has x and y as its input you can write this as doh f so let's say you is equal to f x comma y you are differentiating with respect to t so this is how this expression will come out you will study more about this in your undergraduate syllabus early in your second or third semester of J of sorry of engineering college so now what I'm going to do is I'm going to equate this to zero I'm going to equate this to zero and what I'm going to do is because it is a constant I'm equating it to zero because constant derivative with respect to anything will be zero right so derivative of this with respect to t anything would be zero right now what I'm going to do is I'm going to change my t with an x earlier I didn't want to use x because then you would not get a feel of this term it would become a one actually now I want you to realize that this term is also there so what I'll do I'll put my t as an x okay so this expression here would become doh f by doh x into dx by dx this will become doh f by doh y into dy by dx got it okay so what has happened in this case this will become a one so your dy by dx can be written as minus doh f by doh x divided by doh f by doh y okay so everybody please make a note of this this is the expression for dy by dx for a function which has been implicitly defined this is an expression for dy by dx for a function which has been implicitly defined okay now normally what happens for school exams we don't use this formula okay so I'll write it down use it only for competitive exams use it for competitive exams so how does this work let us take this example itself which I had quoted over here so let's say this is your implicitly defined function and I want to find dy by dx okay now in school what do they say they say that okay differentiate this from left to right and make dy by dx the subject of the formula correct so if you do that it becomes sine square y plus cos xy equal to 1 so let's differentiate both sides with respect to x okay so if you differentiate this it becomes 2 sine y cos y dy by dx what will happen to this term this term will become minus sine minus sine xy into derivative of xy derivative of xv x correct am I right and this will be equal to zero so this is your C that you know the term which you wrote as a C on the yeah this part see okay now let us collect the terms having dy by dx in them so you'll see terms here which is 2 sine y cos y correct from here you will get minus x sine xy okay and this term if you take on the other side it becomes y sine xy y sine x okay so dy by dx will become dy by dx will become y sine xy upon 2 sine y cos y minus x sine xy so this is when you don't use the formula okay so the process is you differentiate from left to right take dy by dx terms common non-dy by dx containing terms send it to the other side and make dy by dx the subject of the formula okay so this is how you would be doing it in the school exam okay for competitive exams what do we do is I'll do it in the second way this is your first way second way if you use the formula that we have discussed right now dy by dx is minus dou f by dou x by dou f by dou y see how the process will become super easy you can do it in one step somebody has a question oh sure dou f by dou x dou f by dou x means you are differentiating by keeping only x as the variable y and constants will be treated as constants so let me ask you this question if you differentiate this with respect to partially with respect to x keeping y is a constant what will happen this time will go for zero what will happen to this term this term will become minus sine y sine xy into derivative of x y now remember your y is a constant so y will come out over here correct hold divided by dou f by dou y dou f by dou y means you have to keep x and constant as constants only y will be treated as a variable so this will become to sine y cos y this will become minus sine xy into x that's it done is this the answer that we had got see one step one step is what is required of course minus and minus will become a plus this is what you will do for your j so one minute sure but please do not use the second formula for school exams you will straight away get no credit for it who knows your parents and all maybe called the other day for meeting and ask what she is doing in the car what he or she is doing the school exams second formula you just have to follow this formula adhan if you're asking what is dou f by dou x then let me tell you dou f by dou x is the derivative of f with respect to x keeping y as a constant of course constants will be constant but y will also be treated as a constant so if you differentiate this keeping y as a constant sine y square or sine square y will be zero because it's a constant cos xy derivative will be minus sine xy into now derivative of x y keeping y as a constant will just give you a y it's like something like 2x so 2 will come out right so same way y will come out at the point no no they'll not be giving you more than two variables but this is extendable this formula is extendable Venkat so if that more than two variables this formula is extendable as you can see du by dt will become dou f by dou x dx by dt dou f by dou y dou y by dt dy by dt plus dou f by dou z into dz by dt yes sir what you want to find out you can find out from there oh okay okay this is something which is not there in your this thing school syllabus or for jee also they don't use it but I'm just telling you in case it may be helpful to you so could you please roll right sir yeah why not yes sir right yes sir one second sure it's a dancer thank you sir then yes sir let's take a question on this yeah see the meaning of dou f by dou x means see when there is a multi-variate function how does your function get affected or what is the change in the function when there is a slight change in the value of x that is called dou f by dou x so what the change in the function when there's a slight change in the value of y that will be called dou f by dou y so do it when it basically says when there is a multi-variate function if you fiddle with one variable how does the function change right that is called dof by dox okay many of derivatives the change in the output when there is a small and nudge in the input budge by nudge right yeah so yeah that is what is dof by dox similarly dof by doi has a similar meaning etc done one minute sir this is easy just to remind you people who join late will stop the class at 7 15 today in the light of your ut is going on yeah thank you so if you see this you can write this as log x squared plus y squared minus 2 tan inverse y by x equal to 0 so you can treat this as your fx comma y okay equal to a constant constant is 0 but it so dy by dx I'll be using my shortcut formula is minus dof by dox by dof by doi okay so dof by dox you what it will be negative of 2x by x square plus y square okay remember you're treating y as a constant here so this will become minus 2 by 1 plus y square by x square into minus y by x square into minus y by x square okay remember y by x derivative when you're doing you'll treat only x as a variable y is a constant whole divided by whole divided by again derivative with respect to y and treating x as a constant will be this minus into 1 by x correct is this fine so few things that we can do directly over here we can make it as 2 plus 2 y so make it as plus 2 y 2 factor you can remove from everywhere if you take x square up if you take x square as a LCM over here this x square gets cancelled with this x square so you end up getting y upon this okay here also we can do the same if you take x square as the LCM over here okay and take this x square on the top that will get cancelled with this x leaving you with the x on the top okay so the numerator term is minus x plus y denominator term is minus y minus x remember the denominators of both numerator and denominator will stand cancelled so this is nothing but x plus y by x minus y which is your RHS and moving on this is basically a twin set of problem which I'm giving you we'll solve only one of them normally in school exams also it appears the question goes like this if under root of 1 minus x square plus under root of 1 minus y square is a x minus y then prove that then prove that dy by dx is given by under root of 1 minus y square by under root of 1 minus x square okay now the various versions of the same problem the next version that you will probably see is this one the next version that you'll see is this one if under root of 1 minus x to the power 6 plus under root of 1 minus y to the power 6 is a x cube minus y cube then prove that dy by dx is going to be x square under root 1 minus y to the power 6 by y square under root 1 minus x to the power 6 okay if you look at a generic case that's what we're going to do now but I'm just uprising you of the different forms of the same problem that you may get if you see the generic case is 1 minus x to the power 2n plus under root of 1 minus y to the power 2n if it is this then prove that dy by dx becomes x to the power n minus 1 under root of 1 minus y to the power 2n upon y to the power n minus 1 under root of 1 minus x to the power 2n let's do the last one let's do this one so what exactly that a doing here sir is just a constant okay but can we do without it also nothing will happen is just a constant very good to see Matthew T is over for you Monday so yesterday this problem if you try to do with brute force really if you try to use your formula also minus dy by dx upon dy by dy you may not get the result very quickly okay you may have to slog you may have to struggle you may have to spend a considerable amount of time to get the output so what do we suggest over here is using some trick substitutions using some trick substitutions okay so what I'll do is I will take x to the power n as let's say sine of theta and I take y to the power n as this is sine of phi okay so what I'm doing is I'm going to take x to the power n as sine of theta and y to the power n as sine of phi it is just for our simplification purpose it has no role to play in the differentiation it is just to simplify that given implicit function okay so basically what I'm doing I'm converting an implicit function into another implicit function okay so now this term over here this term over here will actually become under root of 1 minus sine square theta this term over here will become under root of 1 minus sine square phi okay and this is nothing but sine theta minus sine phi so this is nothing but cos theta this is nothing but cos phi this is nothing but a sine theta minus sine phi okay let us use our transformation formula that we had studied in our class 11th cos a plus cos v formula that's to cos a plus b by 2 into cos a minus b by 2 and this becomes 2 a cos a plus b by 2 into sine a minus b by 2 okay 2 gets cancelled 2 gets cancelled cos term gets cancelled cos term gets cancelled that will give you cot theta minus phi by 2 as a okay that means theta minus phi by 2 is cot inversing take a 2 on the other side okay so this is where we reach out reach up to now what is theta as per the given expression theta is sine inverse of x to the power n and phi is sine inverse y to the power n and this is just a constant you can call it as a c okay now from here we can use our implicit differentiation formula dy by dx is negative doh f by doh x upon doh f by doh y okay so what will it become so negative doh f by doh x will become what 1 by under root of 1 minus square of this term but at the same time we have to use our chain rule so x to the power n derivative is n x to the power n minus 1 now remember the other term is a term in y so that will simply become a zero you don't have to worry about it okay whole divided by doh f by doh y doh f by doh y this will become a zero this will become negative 1 by under root of oh I'm so sorry I think I missed out a 2 1 minus y to the power 2 n into derivative of y with respect to y is this fine now let's simplify these terms so minus here and a minus here will be taken care and n and n will take care of themselves so you end up getting x to the power n minus 1 under root of 1 minus y to the power 2 n divided by y to the power n minus 1 under root of 1 minus x to the power 2 is that fine is this what you wanted to prove is this our right hand side okay now people who are saying done how did you do it was there any other method or did you find longer time to do it people who said done okay okay so this is a very classic problem normally it comes in school exams also but in school exams you'll probably get a lighter version I think the first one you'll get maximum access the second one you'll get but I took a generic case this is a this is a more generic case general case okay let's take another one let's take another one let me put the poll on for this oh sorry Adweta do you want me to go back to the previous page Adweta sorry I missed your message five four three two one please vote everybody most of you have gone with option B let's check see you don't have to start another difference differentiating the function from the word go you can make some small changes into the function for example let me write it like this okay let me square both the sides let me square both the sides so this will give you x square one plus y is equal to y square one plus x okay so this will give you x square y minus x y square plus x square minus y square equal to 0 okay take x y common over here and take x plus y x minus y common over here so that'll give you two expressions over here one is this and the other is that okay now probably x equal to y yeah probably x equal to y is an extraneous factor that has kept into the system okay so this is an extraneous factor because x need not be equal to y right that means the factor which we are interested here is the other one so this is a this is the one where we are interested in okay this is the one which is important to us this probably has come because of the process of squaring as Venkat is rightly pointing out squaring may change the function may introduce extra roots into the system that's why this factor has come into the system okay and if your y is equal to x then divide by dx is going to be one which is not in your options okay so let's not worry about the other factor so let's worry about this guy x y plus x plus y equal to 0 now this is I mean this is actually an explicit function you can express x in terms of y or y in terms of it but let me tell you the formula which I have discussed with you is equally applicable for a cases where you take a bit of effort to separate x and y okay so what I'll do is I'll treat this as fx comma y okay and directly apply my formula dou f by dou x is minus dou f by dou x by dou f by dou y okay so the derivative of this with respect to x keeping y is a constant will just give you a I think why and this will give you a one okay what about the denominator terms denominator term will give you a one plus x okay now hold on this doesn't match with your options but that doesn't mean you have done it wrongly that doesn't mean you have done it incorrectly probably they want you to replace your y completely in terms of x they want you to replace your y completely in terms of x which anyways we can do it from here so we can say why 1 plus x is negative x so why is negative x by 1 plus x okay so place it over here so it becomes negative 1 plus y 1 plus y upon 1 plus x so if you simplify this that clearly leads to the result minus 1 yeah I'm correct minus 1 upon 1 plus x the whole square which is option number b which is option number b now I'm not saying this is the only way to do the problem you could have done it by other ways or as well okay but just be aware that you don't have to start differentiating in the raw form as given to you in the question is that fine any questions here any concerns sir could you go to the left yeah I know thank you sir okay yes sir in this case it was quite easy to determine which was the extraneous roots but then in other cases how exactly do we do it you have to get it clarified from the options you have to see which option yeah which option is basically meeting your requirements okay yes so could you give an example like that there's no ready made example as of now in my mind but I'll do one thing I'll just send you a few questions based on that okay sir I'm unable to put my question so I'll give you my question let's say yeah this is basically a equation of a family of conic passing through origin okay this is basically an example of a family of conic passing through origin it also includes a pair of straight lines by the way okay then then dy by dx is which of the following option a why did I put a bracket on this option is minus x by y option b minus y by x option c y by x option d x by y I'll put the pole on alright at the counter five I'll stop five four three two one go alright so 25 of you have voted so far and 60% of those people say option c okay now very easy question to you know I mean prima facie looks easy but most of the effort probably may go into the simplification because you can see in the answer there is no a there's no b there's no h so to begin with we'll use the formula dy by dx is minus dou f by dou x by dou f by dou y where this is your f this is your f okay so minus dou f by dou x minus dou f by dou x means you're differentiating only with respect to x as a variable so that will be 2 a x this will be 2 h y dou f by dou y will give you 2 h x plus 2 b y okay the factor of 2 can be removed so it becomes negative a x plus h y by h x plus b y now what to do from here on because my answer is in a different format so what I'm going to do is we all know that we can break we can break this term like this so what you can do is we can break this 2 h x y as h x y and h x y so I'll do a x square plus h x y and the other term I'll take it on the other side so minus b y square h x y okay so if you take x common from here you get a x plus h y and if you take a y common from here you get h x plus b y by the way this is b you may look like an h yeah okay so from here we can directly say a x plus h y by h x plus b y of course you can borrow a negative sign that will become y upon x that will become y upon x so this answer that you have got over here is actually y upon x that's nothing but option number c most of you got this correct 60% of you voted for c very good any questions here any any questions here somebody was asking something yes like I mean we also learned let's assume for time being I know I'm complicating it but just for knowledge sake or whatever like let's assume for some time that these are a pair of straight lines okay sir and then okay and let's assume they have been up also like slope is y by x okay that's most of you would have solved it by that method also that's that's a very excellent method of solving it accident I think okay so trippin also has some question trippin you're saying something trippin oh no trippin is same thing same thing yes okay origin then there'll be two slopes yeah so I'm not saying it is differentiable at the origin as you can see it will become zero by zero for yes okay so it is it is non-differentiable at the origin next question I think this is very straightforward you just have to use your formula and your result should be there but if you go by the school method of you know finding dy by dx for the entire expression I mean differentiating it from left to right well you'll have to waste some time collecting your dy by dx terms okay in another 45 seconds we'll stop one please vote please vote only 12 of you voted three minutes is good enough for time for this kind of problem three minutes okay and the poll maximum people have voted for option a a is the most voted for option okay let's check see again if I were you I'll just use my formula dy by dx is negative dou f by dou x by dou f by dou y okay guys are be careful about this negative sign I've seen people missing out on this negative sign as a result they get a negative of the actual answer okay so negative dou f by dou x now dou f by dou x you have to treat x only as a variable everything else should be treated with a constant okay so this will just be 2x e to the power y okay just 2x e to the power y this will be this will be 2 y is like a constant so you can take out so you have e to the power x plus x e to the power x correct all divided by dou f by dou y dou f by dou y the first term will just give you x square e to the power y second term will give you 2x e to the power x okay just do one small thing divide throughout with e to the power x because this doesn't match with any one of the answers so just divide by e to the power x both numerator and denominator so this will give you negative 2x e to the power y minus x minus 2 y x plus 1 upon this will become x square e to the power y minus x and this will become 2x is there any option which says so oh yeah there you go option number 8 so a is absolutely correct much convenient method if you are using this formula but equally cumbersome or you can say you have to collect all your terms with dy by dx collect all your terms without dy by dx and then make dy by dx a subject of the formula but unfortunately you have to follow the same in school you can't do anything about that sir yes sir sir in school just asking could we use the proof you gave like let's say they gave a question and then we use it yes sir they say if you want to use anything else extra I can prove it then use it yeah they say yeah but I've seen them that they did it to marks for that I don't know why they do that but they don't keep their promise yeah seriously I mean some of my students tried proving it law with our rule and they started in a solving it but they were not on it they're not given marks okay so let's let's not try it if at all there's a problem let's not do it yes sir okay let's take easy most of you are able to do it quickly so this is basically why is equal to under root of sine x plus y so y square is equal to sine x plus y now I would basically do this problem by the use of my implicit differentiation I like it like this okay so dy by dx is negative of doh f by doh x so derivative of this is going to be with respect to x is going to be just negative cross x by doh f by doh y so this will give you 2 y this will give you a 0 this will give you a minus 1 so as you can see you can get the answer in straight away two or three steps good enough so could you show the previous slide for one second previous slide for one second yes sir I can't disconnect it huh so that doesn't mean squaring always gives extraneous roots when it it may give extraneous roots