 Hello everyone, myself, Mrs. Mayuri Kangal, Assistant Professor of Mathematics from the Department of Humanities and Sciences, Valchan Institute of Technology, Solapur. Today, we are going to see partial differentiation. The learning outcome is, at the end of this session, the students will be able to solve the examples on partial differentiation. In many fields of science and engineering, we come across with the problems in which we have to find the value of a function which depends upon two or more independent variables. For example, we can see when we have to find the velocity of a board moving in stream, it depends upon the velocity of board, velocity of water, velocity of wind, time, etc. It means to find the value of this function, it depends upon these independent variables. So let us see the need of partial differentiation. The reason for a new type of derivative is that when the input of a function is made up of multiple variables, we want to see how the function changes as we let just one of those variables change while keeping all the others constant. Now what is partial differentiation? A partial derivative of a function of several variables is its derivative with respect to one variable while the other variables are kept constant. In general, if z is a function of two or more independent variables, then partial derivative of z with respect to any one of the independent variable is ordinary derivative of z with respect to that variable treating all the other variables as constant. Now let us see the notations. The symbol used to denote the partial derivatives is denoted here which is called as partially dou or daba. The partial derivatives of a function f of x, y, z and so on with respect to variable x is variously denoted as f to the suffix x or dou f by dou x. Now here this dou is used instead of d in usual df by dx notation to emphasize that this is a partial derivative. This f is a multi-variable function which depends upon two or more independent variables and this x indicates which input variable is changed slightly. Here instead of this dou f by dou x if we have dou f by dou y it indicates that this y at the place of denominator changes slightly and the differentiation is with respect to this variable keeping all the other variables as constant. Now let z equal to f of x, y be a function then partial derivative of z with respect to x treating y as a constant is denoted as f to the suffix x or dou f by dou x or dou z by dou x. Similarly the partial derivative of z with respect to y treating x as constant is denoted as f to the suffix y or dou f by dou y or dou z by dou y. Now we will see the formal definition of partial derivatives of the first order. Let z equals to f of x, y be a function of two independent variables x and y. Then dou z by dou x that is the partial derivative of z with respect to x treating y as constant is given as limit of delta x tends to 0 f of x plus delta x y minus f of x y divided by delta x. Here x is changed and y is kept as constant. Similarly dou z by dou y that is the partial derivative of z with respect to y treating x as constant is defined as limit of delta y tends to 0 f of x y plus delta y minus f of x y divided by delta y. Here y is changed and x is kept constant. Now let us see the use of partial differentiation. The partial differentiation is used to differentiate the functions which have more than one independent variable in them. For example it can be used for finding maxima and minima in the optimization. Now let us see examples. First example, find the first order partial derivatives of z equals to y square plus x cube y. If you observe this z it is of two variables x and y. So z is a function of two variables x y. So we have to find dou z by dou x and dou z by dou y that is the partial derivative of z with respect to x and partial derivative of z with respect to y. Let us go for obtaining dou z by dou x. Here we have to differentiate z partially with respect to x treating y as constant. So we will replace z by the given function dou by dou x of y square plus x cube y. Now to simplify it we will separate the partial differentiation for two terms. So we can write it as dou by dou x of y square plus dou by dou x of x cube y. Now here the partial differentiation is with respect to x so we will treat y as constant. So for the first term which is made up of y only so it is a differentiation will be zero as we know that the derivative of constant is zero. Here y is kept as it is just like a constant and the derivative of x cube that is 3x square is taken here. So we can write it as zero plus 3x square y which gives us the value of partial derivative of z with respect to x as 3x square y. Similarly let us go for obtaining partial derivative of z with respect to y. Here we will treat x as constant. Now dou by dou y of the given function y square plus x cube y. Again we will separate the terms. We can write it as dou by dou y of y square plus dou by dou y of x cube y. Here the differentiation is with respect to y so we will treat x as constant. We know that the derivative of y square is 2y and for the second term x cube is treated as constant and the derivative of y is 1. So we can write the solution as 2y plus x cube. Now pause the video for a minute and write the solution of this question. And the first order partial derivative of z equals to 5xy square minus 2x cube. Now I hope that you all have written the solution. Let us check the solution. Dou z by dou x equals to dou by dou x of z is replaced by 5xy square minus 2x cube. Again separating the terms dou by dou x of 5xy square minus dou by dou x of 2x cube. Here the differentiation is with respect to x so we will treat y as constant. So y square as it is and the derivative of 5x is 5 so 5y square minus sign as it is. The derivative of 2x cube is 6x square so we get dou z by dou x as 5y square minus 6x square. Similarly dou z by dou y which is obtained by differentiating z partially with respect to y treating x as constant. Let's separate the terms here the differentiation is with respect to y and x is treated as constant. So 5x as it is and the derivative of y square is 2y which gives us 10xy minus sign as it is. Here 2x cube is there and the differentiation is with respect to y so its derivative will be 0 as it is treated as constant. So we get 10xy minus 0 which gives us 10xy. Now let us go for the second example. Find the first order partial derivatives of u equals to x square y cube z at the point 1 to 1. Here u is expressed as a function of 3 variables xyz so u is a function of xyz and we have to find the first order partial derivatives. So we have to find dou u by dou x dou u by dou y and dou u by dou z at the point 1 to 1. Now differentiating u partially with respect to x treating y and z as constant so dou by dou x of x square y cube z here y and z as treated as constant so y cube z remains as it is and the derivative of x square is 2x so we get 2xy cube z. Similarly dou u by dou y differentiating u partially with respect to y here we treat xz as constant in this term dou by dou y of x square y cube z x square z are kept as it is and the derivative of y cube is 3y square which gives us the value of dou u by dou y as 3x square y square z. Similarly dou u by dou z the differentiating u partially with respect to z treating x and y as constant gives us dou by dou z of x square y cube z here xy are treated as constant so x square y cube as it is and the derivative of z is 1 which gives us the value of dou u by dou z as x square y cube. Now let us find the values of these partial derivatives at the point 1 to 1 now dou u by dou x is 2xy cube z dou u by dou y is 3x square y square z and dou u by dou z is x square y cube therefore dou u by dou x at the point 1 to 1 is obtained by substituting x as 1 y as 2 and z as 1 so 2 as it is x is replaced by 1 y is replaced by 2 and z is replaced by 1 which gives us 2 into 1 into 2 cube into 1 we know that 2 cube is 8 into 2 that is 16 so the dou u by dou x at the point 1 to 1 is 16. Now dou u by dou y at the point 1 to 1 dou u by dou y is 3x square y square z again x is replaced by 1 y is replaced by 2 and z is replaced by 1 which gives us the value of dou u by dou y at the point 1 to 1 as 3 into 1 square into 2 square into 1 which gives us 2 square 4 into 3 that is 12 and dou u by dou z at the point 1 to 1 is x square y cube which in which x is replaced by 1 and y is replaced by 2 which gives us 1 square into 2 cube 2 cube is 8 into 1 gives us the value of dou u by dou z at the point 1 to 1 as 8 thank you.