Stony Brook MAT 125 Spring 2015
Lecture 12: Implicit differentiation
March 11, 2015

Start   we had the chain rule the chain rule drives some of you crazy so before we do differentiation were just going to practice the chain rule for a few minutes so remember to do the derivative of the chain rule you have 2 functions f of g of x you could actually have 3 functions r 20, you can mess this or clean this how you want
0:31but the principle all you need is 2 functions you have an outside you have an inside in order to do the derivative you do the outside function you leave the inside alone times the derivative of the inside function so another words you take the derivative of the outer funtion you dont do anything to the inner function times the derivative of the inside function we i know a lot of you, the next webassign will have a lot of chain rule function
1:03not due for a couple weeks so better work on it dont do it till you feel comfortable so heres the type that trips people up a lot something like that so take a second and see if you can figure it out this is puzzling lots of you theres 3 layers to this
1:31because if you rewrote this this is really cos of 10x cubed so you have 3 things going on, so always ask youreself what is happening to x first i take x and multiply it by 10 second i take 10x and take the cos and then third i take cos of the 10x and cube it you got three layers, f g and a h so how would we do the derivative
2:011 layer at a time but first im just doing the derivative of the something cubed so its 3 and the something squared now im going to fill in the something that doesnt change, thats just cos10x then i do the derivative of cos which is minus sin of x notice i still have the 10x but that hasnt changed
2:30notice their isnt any squaring anymore its just the derivative of cos hen i do the derivative of 10x which is 10 much easier to watch me do it right good so got the ide lets do another one almost the same but slightly different this time we have 3 layers again just like the last one
3:06we have something to the 4th we have sec and pix so you say what am i doing to x, first i multiply it by pi then i take pix and take the sec of it then i take the sec of pix and raise it to the 4th 3 levels just like last time so the derivative, first i have to do the derivative of whatever to the 4th this thing to the 3rd
3:35its just 4 sec pi x cubed now take the derivative of sec pi x this is where people mess up thats sec pix tangent pix then you have to do the derivative of pi x which is pi who got that, that makes me feel better
4:01lets try one more last one so this is another 3 layer one again, but this time i use the cubed root
4:31this is cot of x squared plus 1 to the 1/3, because cubed root is to the 1/3 power so first i have to take the derivative of something to the 1/3 which is 1/3 cot of x squared plus 1 so you dont do anything to the inside now 1/3 this minus 2/3 because 1/3 minus 1 is -2/3
5:01this is where you guys will trip up, youll do 1/3 - 1 and youll do that wrong okay so practice make sure you get the hang of that then you have the derivative of cot which is negative cos squared and you dont do anything to the inside and then times the derivative o x squared plus 1 which is 2x howd we do on this one?
5:30not as good not as good okay so we still have quite got it down yet but were getting there how much time do i have, i have 5 more minutes how about
6:01lets take the derivative of that this is going to require the product rule because we have 2 things going on, we have 5x say to yourself whats going on with x you take the cosx and do the e of that and thats all you can do tp that step this x you multiply by 5 so these are really two different things that are happening to x so thats how you know its a product rule because its two different functions
6:30okay so ill repeat that x you take the cos of e and then take the e to the cos so thats 1 thing being done to x and the separate thing is multiplied by 5 thats why you need the product rule you will almost always need a chain rule to do the derivative because the only time you wont need chain rule is when you just have a function of x so if you have sin of x or tan of x or log of x then you wont need the chain rule but as soon as theres a number in there, sin of 5x
7:04or the natural log of x squared plus 1 or anything like that now you can do the chain rule product rule first times the derivative of the second plus 2nd times the derivative of the first you have the first function now whats the derivative would be of cosx well we have an outside function and an inside function, the outside is e to the so youd just do e to the whatevers up there derivative of e to the x is e to the x
7:32times the derivative of cosx which is negative sinx plus reverse the derivative of 5x is 5 e to the cosx if you wanted to simplify this you could pull out e to the cosx, by the way you could also pull out a 5
8:03but you could pull out an e to cosx when you have a product rule with an e function in it, with an exponential function in it you will always be able to factor the e term out of the power rule after you get the derivative that term will survive in all of the terms if we take it out 1st derivative 2nd derivative 3rd derivative thats going to be important because when you set this equal to 0 youll know youll have this term youre setting equal to 0
8:31and e to the anything is never 0 so remmeber that youll always pull this out we have time for 1 more, sure
9:07heres a nice sophisticated this is chain rule again cosx+1 over sinx plus 1 minus 1
9:31all raised to the 1/2 gonna need some space for this one so you say to yourself whats happening to x well first i take the cos of it and add 1 and in the second function i take the sin and subtract 1 his is going to be a quotient rule then i take that whole thing and raise it to a half
10:03first i do the derivative if the outside which is 1/2 whole mess and the minus 1/2 and i dont touch whats inside now i have to take the derivative of that thats going to require the quotient rule lodhi, hidlo minus lo squared
10:30so its a low function sinx-1 times the derivative of the high function which is minus sinx minus reverse leave the top function alone, cosx+1 now we do the derivative of the bottom function which is cosx all over sinx-1 squared you can simplify that if you wanted
11:02why would you want to siplify that well if i was really a horrible person id ask you to take the second derivative you really wouldnt want to do that you might need to solve it for 0 otherwise you would just leave it alone, if this was a question like find the equation of a tangent line of pi/6 you would think 1pi/6 right away cause you guys all have to the sin an cos of pi/6
11:30but if you dont have to do anything else you stop this however will multiple nicely, minus sin squared and a minus cos squared and have 1 or minus 1 so it would actually simplify a lot but you dont want to have to do that, yes what happened to a 1/2 well its a half this whole thing is a minus 1/2 and you stop and now youre just doing the inside the otter funtion is the 1/2 power
12:01so its 1/2 inside function to minus 1/2 then you only have to take the derivative on whats the inside how are we doing on these you have a couple weeks, who go them all right now were going learn somehting called implicate depreciation kind of the chain rule
12:33theyre related its not kind of its related alright so far all the derivatives we have have been in the order of y equals soemthing you have a function and you got y isolated everything on the other side is a different period, x t or whatever what do you do when you have a function and you cannot
13:00seperate the variables so if you had something like and you have to take the derivative of that, and that looks pretty easy but the problem is is not, find dy/dx the problem is its not y= something of x and you cant easily solve for y probably if you do some really complex algebra you can isolate y
13:31but otherwise its hard to do so the technique call them implicate depreciation who teach you to define dy/dx or any derivative without isolating y then you get the answer whats called implicitly because it will be in terms of x and y its implicate because you dont actually know what x and y are so what do i mean so what does it mean when you have dy/dx
14:00dy/dx is really the slope its saying how does y change in respect x, another words how does y change and x change if i had if i had y= a function of t dy/dt would be how would y change as t changes if i had y as a function of w i would do dy dw, i would want to know how y changes when w changes but what if i had y=ft and wanted to find out the change in dx
14:35well now the problem happens over here, i dont know what to do because t has to be in terms of x somehow but that point implicate depreciation is going to be 4 so heres what you do every term youre going to take the derivative of is multiplied by d of that variable dx in this case what do i mean i take the derivative of x squared which is 2x
15:01dx, dx because thats, how does that change as x changes becasue you could think of this of x,x i know that sounds weird that would have to be 1 but ignore it then i take the derivative of 3y squared and i say how does y change as x xhanges then i take the derivative of 8x then i minus 8 how does x changes as x changes nothing special happens
15:31then the derivative of 4y cubed which si 12y squared dy/dx so again were trying to find how y changes when x changes so y is a function of x so i do the chain rule, so the otter function would be 6y and then inner function would be how does y change with the effect of x which is right because you want it at the end when i do the derivative of x squared its the otter function 2x
16:00the inner function is just x so there is no inner function thats why that derivative is just 1 and we end up ignoring it you dont actually have to write that term if you dont want to but you should know its sort of there in other words this is 2x dont worry well do a few more of them so its 2x
16:306y dy/dx minus 8, 12y squared sy/sx and now comes the part where you all have problems with the algebra so you would have to isolate dy/edx this is something we did last semester to our logarithms its the same technique that you use for isolating dy/dx theres 3 steps the steps are groups
17:00factor, divide these are going to be the steps, so you look at your equation move everything that contains dy/dx on one side and everything that does contain dy/dx on the other side so on the left side ill put whatever has dy/dx 6y dy/dx plus 12y squared dy/dx and on the other side i put everything that does not contain dy/dx
17:30i got 2x and i got 8 8-2x so the 2 dy/dx terms stay on the left side the two terms without dy/dx go over to the other side then i factor out dy/dx so again i grouped
18:04i factor and now i divide so i get dy/dx alone equals 8-2x over 6x plus 12y squared and that is dy/dx so notice its got an x and a y in it thats why its called implicate, we dont actually know how to find y because we have x and ys mushed together here
18:32so ill do a couple more examples, dont worry i wont just leave you with 1 why is it 6x?
it should be 6y, just because im old
19:03lets do another one to make sure you guys have the hang of it okay so so whats going on remember you have y as a function of x so what you do the derivative of that its a chain rule you have whatever that function is and you multiply it by the derivative of y okay when you have x wed think of it as x as a function of x but that doesnt really mean anything
19:30so you would write dx/dx but dx/dx is 1` so we dont really follow the rule right lets try another example something like that you have 8x to the fourth minus 10y to the 5 plus 6x cubed equals
20:0011y squared plus 4 were going to take the derivative of that implicately how would you know to use the implicate depreciation youll have xs and yx or whatever youll have more than one variable in a question and you wont be able to isolate a variable cause by the way because if i didnt have this 11y squared term you could just move everything over to the other side take the fifth root of y and now you got y by its self
20:30the problem is you got 10y to the 5th plus 11y squared you cant use the qiuadratic formula if thats a fifth power by the way you wouldnt necessarily want to if i gave you you could do the quadratic formula if you take the derivative of that but ou wouldnt want to do that but its much easier to do it implicitly
21:03the idea is this is what you do when you cant isolate y so lets practice, first you take the derivative of each term the derivative of 8x to the 4th is 32x cubed times the derivative of x which is dx/dx which we cross out so were going to practice writing it a couple more times take the derivative of 10y to th e 5th which is 50 y to the 4th
21:31times the derivative of y then i got 6x cubed which is 18x squared times derivative of x equals the derivative of 11y squared is 22y derivative of y plus 0
22:11group factor and divide so group everything with a y term goes on one side everything without a term goes on the other side cross out the dx/dx so i got this has a dy/dx and this has a dy/dx lets put them both on the same side
22:34and youre left with 32x cubed plus 18x squared equals 22y dy/dx +50y to the 4 dy/dx everything with a dy/dx goes to the right
23:01everything without dy/dx goes to the left factor dy/dx out so you got the group, you got the factor so 32x cubed plus18x squared equals dy/dx times 22y plus 50y to the 4
23:30then divide last step and you get 32x to the 3 plus 18x squared over 22y plus 50y to the 4 thats not so bad youll get the hang of it well practice another one right now before i get nasty
24:26there we go
24:30y 8 minus 6y to the 5 plus11x,7 +32x squared = 5x+4 so you can move the x terms to the right right away you get there faster but were still stuck with dy terms derivative of y to the 8 is 8y to the 7 dy/dx the derivative of 6y to the 5
25:01is 30y to the 4th dy/dx by the way some of you like y prime cuase you learn this in high school and your teacher used prime your going to want to learn to do it this way the derivative of 11x to the 7 77x to the 6 now theres a dx/dx there but im not going to write it then there is derivative of 32x squared which is 64x
25:30equals 5 plus 0, so far so good lets do it part terms we have 8y to the 7 dy dx minus 30y to the 4 dy/dx equals 5 minus 77x to the 6 minus 64x
26:00so far so good factor out dy/dx 8y to the 7 minus 30y to the 4th equals the mess on the right side and divide im hearing very ahppy nosies this means you guys are ready for the harder problem
26:30bring it on you get 5-77x to the 6 minus 64x over 8y to the y minus 30y to the 4 okay how can i make this harder?
27:13so how could i make this harder? well how bout i give you one of these find dy/dx
27:31youre going to need the product rule so lets take the derivative so first we have to take the derivative of 5x cubed, thats 15x squared times the derivative of x and the derivative of x is just 1 so ignore that, now for the fun part product rule, so we have 2 functions we have 4x squared and we have y squared so first lets leave the four x squared alone
28:00times the derivative of y squared which is 2y dy/dx plus reverse because this is product rule derivative of 4x squared is 8x times the derivative of x which is 1 times y squared thats the middle part plus the derivative of 6y cubed 18y squared
28:30dy/dx equals 0 now suppose this problem was find the tangent line of an equation like this at x=1 y= whatever you immedaitely plug in you dont want to do the algebra of isolating dy/dx thats a lot of work theres a chance youll mess it up algebra can be a bit tricky but however we say find dy/dx now you have to do all the steps lets do this again, make sure you understood what i did
29:00the derivative of 5x cubed is 15x squared and 4x squared y squared you have two functions here you have 4x squared, y squared s its four x squared times y squared which is 2dy/dx plus derivative of 4x squared which is 8x times the derivative of y squared taking the derivative of 6y cubed is 18y squared dy/dx derovative of 1 is just 0 so before we do our technique which have to do some simplifying so 15x cubed
29:34and you get lets see 8x squaredy dy/dx plus 8xy squared plus 18y squared dy/dx now group factor divide so now lets put everything with dy/dx and leave it on the left side
30:01more the other term to the right side so i get 8x squared y dy/dx plus 18y squared dy/dx equals -15x squared minus 8xy squared
30:31factor out dy/dx and divide
31:00-15x squared -8xy squared over 8x squared y plus 18y squared lets do another one how else would i make this really hard, yes why isnt ther a dy term
31:31when i took the derivative of y squared i got 2 dy/dx now put in the product rule times the derivative of x which is 1 the derivative of the 4x squared doesnt contain any ys so thats why theres no dy/dx remember im doing the derivative of 2 different functions
32:05so one more time, if i just had the derivative of 4x squared y squared, its the derivative of 4x squared timesthe derivative of y squared dy/dx plus y squared times the derivative of 4x theres no dy/dx because its just 8x times the derivative of x which is 1
32:31this is tricky you have to do it a few times thats why ill put more stuff online after you do 3 or 4 thousand of them youll get good at them could i make this more annoying oh sure
33:04lets find dy/dx the derivative of sin, the outside function is sin you get cosxy, you get nothing times the derivative of xy thats going to require the product rule x times the derivative of y which is dy/dx
33:32plus y times the derivative of x thats your left side the right side is 1 plus dy/dx that wasnt so bad were not done, oh no we have to isolate dy/dx how do we do that you dont divide like cos, you distribute
34:01so first you take this x cos x dy/dx plus y cosx y equals 1 plus dy dx now the same as before move the dy/dx terms to one side and the non dy/dx to the other
34:30so i put y/dx to the left i get x cosxy dy/dx minus dy/dx equals 1-y cosxy next pull out dy/dx
35:04xcoxy minus 1 equals 1-cosxy and divide you get dy/dx 1-y cosxy
35:30over xcosxy minus 1, normally i could ask for the second derivative but i wouldnt do that remember what i told you the very first class the problem with clac isnt the derivative part its the calculus part this first step isnt so rough its the remaining part that gets you through that messes you up so thats what you practice, remember what i said ill put some stuff on blackboard