# First Order Partial Derivatives – Problem 1 – Partial Differentiation – Engineering Mathematics 1

Do subscribe to Ekeeda channel and press bell icon to get updates about latest Engineering HSC and IIT-JEE Mains and Advanced videos.Hi Students, so after understanding how to find out the partial derivative of first-order and second-order let’s start with a numerical based on partial derivative of first order so your i am gonna cover numerical where B I want to find out the partial derivative first-order evaluate the given numerical. So here u is given as 1 minus 2 XY plus y squared whole raise to minus up and you have to do that X then u by Del X minus y del u by Del y is equal to y square u cube so to get the answer it’s simple because here guys then u by Del X and del u by Del y these two terms you know that this is partial derivative of first order so what I’m gonna do is i am gonna differentiate is U with respect to X partially and then we put that value over here and then we’ll differentially U with respect to Y partial we’ll put that value over here by multiplying with X&Y and subtracting will get the answer so let’s start so first of all the partial differential DC U with respect to X now sure we get dou u by dou X or it is also called a net u by Del X so there’s there are many pronounciation to this some people call it at D by you by Debb IX also so I call it as del u by Del X now it is equal to so we’ll find out the differentiation of U with respect to X so just now in the previous video we have covered how to find out the partial derivative of a function with respect to one variable so as per the rule me keep the second variable as sponsor or p3 second variable X constant so since I’m differentiating with respect to X I am going to treat Y as constant so you’re if I differentiate this with respect to X I’ll get negative half 1 minus 2 XY plus y square whole raised to negative 3 upon 2 and then maybe here this function we will differentiate it with respect to X again so differentiation of 1 is 0 for this negative 2xy this -2 and y both are constant because as you are different here with respect to X we are gonna consider Y is constant and then derivative of X with respect to X is 1 then difference you know Y square with respect to X is 0 so you have we got negative 2 we’re now guys we are gonna multiply this value with X so here and say X dou u by dou X or X del u by Del X now here 2 n 2 will be cancelled negative and negative will become positive Y and since I’m multiplying with X that will become X Y and this term will remain as it is which is now I call this as equation number one now similarly I find the second term and call this as equation number two and then we’ll subtract these two terms of these two equations so here I’ll say to find out it then u by Del X del Y I’ll say partial differentiating u with respect to Y so you will get del u by Del y equal to which is negative half 1 minus 2xy plus y square whole raised to negative 3 by 2 and again derivative of this function with respect to Y partially so that time we’ll get 0 for this one this is negative 2 X is constant there 80 of Y with respect to Y is 1 and for y square we will get positive 2 1 now again I can take two common from these two terms I will cancel it with this tool and we’ll send this negative sign inside so this negative X will become positive X and this positive Y will become negative I so here I get X minus 1 into this term nowadays we also have to multiply this term with Y and if I multiply it Y with Y on the left hand side I get extra Y on the right hand side so therefore we’ll get why del u by del y equal to Y into X minus y now when I call this as equation number two so we got value of both the terms in equation one and two now we’ll subtract it to get the answer so subtracting these two equations so you will get left hand side will be X del u by Del X minus y del u by Del y equal two now if I’ll subtract these two terms on the right hand side this is the second equation this one is the first equation from first and second equation we can take this bracket common so we will get 1 minus 2 X Y so this service common and now let’s write down the different terms in a bracket so from here we’ll get X Y so in the bracket I’ll write down that X 1 and from here since we are subtracting this so I will be having negative sign this term which is Y into X minus 1 now let’s solve this guys and let’s check whether we are getting answer or not so this will be 1 minus now in the bracket will get x1 by multiplying with negative I will get negative x 1 and plus y square now we have this X 1 X 1 will be canceled and we’ll get Y square into this now let’s come here should you’ll see here in the question the value of this term is given as u now your we have the same function the only difference is the power so I’ll rewrite the power a and I’ll do some adjustment like this so from here we got only Y Square after cancelling these two terms and from here I’ll read a disturb as so what I did here is I’ve kept negative 1 by 2 inside and taken 3 outside and because of this this bracket has become you because you can observe that this value is nothing but you so guys that becomes u cube and we will get answer as Y square into YouTube and guys this is nothing but the required value so yeah we got the answer so I’m sure that you understood the give it explanation and you understood how I got or have a proved a given result so guys keep watching the videos if you like 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@t

very well explanation sir shake hand from you <3 nd sounds too good with clear voice and pretty heart

What happened to + y square

bhai aram se padha bahut jaldi jaldi padha rha hai thodi hindi bol english jaydha samjhe nhi pata

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