 Hello everyone! Myself, Mrs. Mayuri Kangle, Assistant Professor of Mathematics from the Department of Humanities and Sciences, Valtran Institute of Technology, Soolapur. 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. Let us see the type 1 examples. Let z equals to f of u and u equals to phi of x, y so that z is a function of u and u itself is a function of two independent variables x and y. See here, z is a function of single variable u and u is a function of two variables x and y. In such cases, z will be a function of x and y. For example, let z equals to tan u and u equals to x square plus y square then z is equal to tan of x square plus y square where u is replaced by x square plus y square. In such case, we can see that z is a function of two variables x and y. Now, let us see the partial differentiation of type 1 examples. Let z equals to f of u and u equals to phi of x, y possesses first order partial derivatives. See here, z is a function of single variable u and u is a function of two variables x and y. In such cases, we have seen that z is a function of x and y. So, to differentiate z with respect to x and y, we have to go for partial differentiation from which we can obtain dou z by dou x and dou z by dou y. Here, z is a function of single variable u so we have to go for ordinary differentiation d by du and u is a function of two variables x and y. So, we have to go for partial differentiation dou by dou x and dou by dou y by using which we can find out the two partial derivatives dou z by dou x and dou z by dou y. To obtain dou z by dou x, we will multiply dz by du and dou u by dou x. So, we get dou z by dou x equals to dz by du into dou u by dou x. Similarly, to obtain dou z by dou y, we multiply dz by du and dou u by dou y. So, we get dou z by dou y equals to dz by du into dou u by dou y. For example, z equals to sin u and u equals to phi of x y. Then, dou z by dou x equals to dz by du into dou u by dou x. Now, z equals to sin u. So, dz by du is cos u. So, we get dou z by dou x equals to cos u into dou u by dou x. Similarly, dou z by dou y equals to dz by du into dou u by dou y. So, we get dou z by dou y equals to cos u into dou u by dou y. Now, let us see examples over it. First example, if z equals to e raised to u and u equals to 4x minus phi y, then find dou z by dou x and dou z by dou y. Here, z equals to e raised to u. That is, z is a function of single variable u. So, we can find dz by du equals to derivative of z with respect to u. That is, e raised to u. We will call it as equation number 1. And u is given as 4x minus phi y, which shows that u is a function of two variables x and y. So, differentiate u partially with respect to x as well as y. When we perform the partial differentiation of u with respect to x, we treat y as constant. So, the derivative of minus phi y will be 0 and the derivative of 4x gives us 4. So, dou u by dou x equals to 4 and dou u by dou y. That is, the partial differentiation of u with respect to y treating x as constant. So, the derivative of 4x will be 0 and the derivative of minus phi y with respect to y is minus 5. So, we get dou u by dou x equals to 4 and dou u by dou y equals to minus 5. We will call it as equation number 2. Now, we know that if z is a function of u and u is a function of two variables x and y, then dou z by dou x equals to dz by du into dou u by dou x and dou z by dou y equals to dz by du into dou u by dou y. Using equation 1 and 2, we will replace dz by du, dou u by dou x and dou u by dou y. So, we get dou z by dou x equals to e raised to u into 4. That is, 4 e raised to u and dou z by dou y equals to e raised to u into minus 5. That is, minus 5 e raised to u. Now, to get the solution in terms of x and y, we will put u equals to 4x minus 5y. We get dou z by dou x equals to 4 e raised to 4x minus 5y and dou z by dou y equals to minus 5 e raised to 4x minus 5y. Now, pause the video for a minute and write down the solution of this question. The question is, if z is equals to f of u and u equals to g of a, b, c, then find dou z by dou a, dou z by dou b and dou z by dou c. I hope you all have written the solution. Let us check the solution. Here, z is given as a function of single variable u and u is given as a function of 3 variables a, b, c. See here, z is a function of single variable u. So, we can find out dz by du and u is a function of 3 variables a, b and c. So, we can get dou u by dou a, dou u by dou b and dou u by dou c. Now, this z is a function of 3 variables a, b, c. So, we can find out dou z by dou a. So, dou z by dou a equals to dz by du into dou u by dou a. Similarly, dou z by dou b equals to dz by du into dou u by dou b and dou z by dou c equals to dz by du into dou u by dou c. Now, let us go for the second example. If u equals to log t and t equals to x cube plus y cube minus x square y minus x y square, then prove that x into dou u by dou x plus y into dou u by dou y equals to 3. Given that u is equals to log t, here u is a function of single variable t. So, we can get du by dt by differentiating log t with respect to t. So, du by dt equals to 1 upon t. We will call it as equation number 1 and t is equals to x cube plus y cube minus x square y minus x y square. Here, t is a function of 2 variables x and y. Differentiating this t partially with respect to x treating y as constant gives us dou t by dou x equals to differentiation of x cube is 3 x square y is treated as constant. So, the differentiation of y cube will be 0 minus x square y is there. It is a differentiation with respect to x treating y as constant will be minus 2 x y and here for minus x y square the differentiation of x is 1. So, we get minus y square only. So, dou t by dou x equals to 3 x square minus 2 x y minus y square. Now differentiating t partially with respect to y treating x as constant. So, the derivative of x cube will be 0. The differentiation of y cube is 3 y square x is treated as constant. So, minus x square remains as it is and the differentiation of y is 1. So, we get minus x square again for this term minus x y square minus x is treated as constant and the differentiation of y square is 2 y. So, we get dou t by dou y equals to 3 y square minus x square minus 2 x y. We will call it as combinedly as equation number 2. Here u is equals to f of t and t equals to g of x y. Then dou u by dou x equals to du by dt into dou t by dou x and dou u by dou y equals to du by dt into dou t by dou y. To obtain the partial differentiation of u with respect to x and y we will use the equation number 1 and 2. So, we get dou u by dou x equals to 1 upon t into bracket 3 x square minus 2 x y minus y square and dou u by dou y equals to 1 upon t into the bracket 3 y square minus x square minus 2 x y. Now, we have to prove that x into dou u by dou x plus y into dou u by dou y equals to 3. Now, to prove this let us consider the left hand side. So, LHS is equals to x into dou u by dou x plus y into dou u by dou y. So, we will multiply dou u by dou x by x and dou u by dou y by y. So, we get x into 1 upon t into the bracket 3 x square minus 2 x y minus y square plus y into 1 upon t into the bracket 3 y square minus x square minus 2 x y. Now, if you observe the 2 terms 1 upon t is common in both the terms. So, let us take 1 upon t common into the bracket 3 x square into x gives us 3 x cube minus x into 2 x y gives us minus 2 x square y minus x into y square gives us x y square plus this y is multiplied to the bracket. So, 3 y square minus x square y minus 2 x y square that is equals to 1 upon t into the bracket 3 x cube plus 3 y cube as it is if you see we have the 2 terms with x square y that is minus 2 x square y and 1 more minus x square y which gives us minus 3 x square y. Similarly, for the term x y square here we have minus x y square minus 2 x y square which combinedly gives us minus 3 x y square. If you observe this bracket we have 3 common in each term. So, taking 3 common which gives us 3 upon t into the bracket x cube plus y cube minus x square y minus x y square as we know that x cube plus y cube minus x square y minus x y square is t we can replace this bracket by t. So, we get 3 upon t into t here t get cancelled and we get the solution as 3 which is the required RHS the right hand side of the equation. Thank you.