 Hello everyone, myself A.S. Falmari, Assistant Professor, Department of Humanities and Sciences, Valshant Institute of Technology, Solaapur. In this video, we are going to learn about Jacobians. The learning outcome of this video is, at the end of this session, students will be able to find the Jacobian of given functions. Let us start this lecture with the definition of a Jacobian. If u, v are functions of two independent variables x, y, then the determinant dou u by dou x, dou u by dou y, dou v by dou x, dou v by dou y is called Jacobian of the functions u, v with respect to independent variables x, y and this Jacobian is denoted by these symbols and this symbol can be read as Jacobian of the functions u, v with respect to independent variables x, y that is the Jacobian of any two functions is nothing but it is a determinant. In that determinant, the first row contains the partial derivatives of the first function with respect to all the independent variables and second row contains the partial derivative of the next function with respect to all the independent variables. Now, just we have defined Jacobian for two functions in two independent variables. We can extend the definition for three functions in three independent variables as if u, v, w are the functions of x, y, z then the Jacobian of u, v, w with respect to x, y, z is denoted by this symbol and it is defined as the three by three determinant where the first row contains the partial derivatives of u with respect to x, y, z. Second row contains the partial derivative of v with respect to x, y, z and the last row contains the partial derivative of w with respect to x, y, z. We can similarly define the Jacobian of n functions in n independent variables. Now, pause this video and answer this question if u equal to x into cos y, v equal to y into sin x then find dou u by dou x, dou u by dou y, dou v by dou x, dou v by dou y. I hope that all of you have written the answer. The problem is to find out all the partial derivatives of these two given functions. Let us differentiate first function u partial with respect to x treating y constant. As y constant we can write cos y as it is and the derivative of x with respect to x is 1. Again differentiating u partial with respect to y treating x constant as x is constant we can write x as it is and the derivative of cos y is minus sin y. Therefore, the derivative becomes minus x into sin y. Similarly, differentiating v partial with respect to x treating y constant as y constant we can write y as it is and the derivative of sin x is cos x. And finally, differentiating v partial with respect to y treating x constant as x is constant we can write down sin x as it is and the derivative of y with respect to y is 1. Let us consider the first example. If u equal to x plus y upon 1 minus x into y and v equal to tan inverse x plus tan inverse y find Jacobian of u v with respect to x y solution. Now, here we have provided two functions in two independent variables therefore, Jacobian of these two functions with respect to x y is defined by the 2 by 2 determinant in which the first row contains the partial derivatives of u with respect to x y and the second row contains the partial derivative of next function v with respect to x and y. Now, in order to find out this determinant we must know all these entries. Let us find out all these partial derivatives first differentiating u partial with respect to x treating y constant. Now, here we can observe that u is provided in terms of a rational function where the numerator and denominator both contains functions of x therefore, in order to differentiate it we have to use u by v rule. The rule is what? Writing denominator 1 minus x into y as it is into the derivative of the numerator x plus y with respect to x treating y constant is 1 minus the numerator x plus y as it is into the derivative of the denominator 1 minus x into y with respect to x treating y constant is minus y divided by square of the denominator that is 1 minus x into y bracket square it is equal to multiplying by 1 to this bracket results in the bracket itself 1 minus x into y and this minus minus becomes plus now simply multiply this bracket by y we get plus x y plus y square. Now, here we can see that these two terms will be removed and final answer we get dou u by dou x is equal to 1 plus y square upon 1 minus x into y bracket square. Now, again differentiating u partially with respect to y treating x constant now again y is involved in the numerator and denominator also let us use again one more time u upon v rule let us write down denominator 1 minus x into y as it is now the derivative of the numerator x plus y with respect to y is 1 minus writing the numerator x plus y as it is into the derivative of the denominator 1 minus x into y with respect to y is minus x divided by 1 minus x into y bracket square it is equal to multiplying by 1 to this bracket is the bracket itself 1 minus x into y and this minus minus becomes plus multiplying by x to this bracket we get it as plus x square plus x y now here again minus x y and plus x y will be removed the final answer is 1 plus x square upon 1 minus x into y bracket square. Now, let us find out the partial derivative of v with respect to x treating y constant is a function of y which is constant therefore, its derivative is 0 and the derivative of tan inverse x is 1 upon 1 plus x square now again differentiating v partially with respect to y treating x constant the derivative of first term tan inverse x is 0 and the derivative of tan inverse y with respect to y is 1 upon 1 plus y square now substituting all these four derivatives in this determinant we get this expression now evaluating this determinant 1 plus y square upon 1 minus x into y bracket square into 1 upon 1 plus y square minus this product 1 plus x square upon 1 minus x into y square into 1 upon 1 plus x square this 1 plus y square 1 plus y square will be removed here also 1 plus x square 1 plus x square is removed and we get this expression as 1 upon 1 minus x into y bracket square minus 1 upon 1 minus x into y bracket square and which is equal to 0. Let us consider one more example if x equal to r sin theta cos phi y equal to r sin theta sin phi z equal to r cos theta show that Jacobian of x y z with respect to r theta phi is equal to r square sin theta now here we want to find out the Jacobian of three functions with three independent variables so we know that this Jacobian is provided in terms of the determinant of order 3 by 3 the entries of the determinants are the partial derivatives so in order to find the Jacobian let us differentiate x partially with respect to r treating remaining variables constant. Here sin theta cos phi are constant and the derivative of r with respect to r is 1. Now differentiating x with respect to theta treating remaining variables constant that is r and cos phi are constant derivative of sin theta is cos theta and finally differentiating x partial with respect to phi treating remaining constant that is r and sin theta constant and the derivative of cos phi is minus sin phi. Now let us consider the next function y. Now differentiating y partially with respect to r treating remaining variables constant. Now the derivative of r is 1 and these are the constant we can write them as it is sin theta into sin phi. Now differentiating y partially with respect to theta treating remaining variables constant. Now we can write r as it is and sin phi as it is the derivative of sin theta is cos theta. And finally differentiating y partially with respect to phi treating r and theta constant as r theta constant we can write r sin theta as it is and the derivative of sin phi is cos phi. And now the final function differentiating z partially with respect to r treating remaining variables constant here z equal to r cos theta cos theta is constant derivative of r is 1. Now differentiating z with respect to theta treating r and phi constant here phi is not involved r is involved it is constant and the derivative of cos theta is minus sin theta. Therefore the total derivative is minus r into sin theta as it will not includes phi partial derivative of z with respect to phi is 0. Now the definition of Jacobian of three functions in three independent variables is these three by three determinant. Now substituting all the partial derivatives in this determinant we will get this expression. Now from the second column and in the third column we can observe that it contains a common factor r therefore we can take r common from these two columns. So outside we get it as r square and in the last column also we can observe that sin theta is also present here 0 can be written as 0 into sin theta. Therefore taking sin theta common we get outside r square into sin theta. Now the remaining determinant is sin theta cos phi cos theta cos phi minus sin phi sin theta sin phi cos theta sin phi cos phi and cos theta minus sin theta 0. Now simplifying this determinant along the last row now we get this r square sin theta as it is cos theta into deleting the last row and the first column we get this 2 by 2 determinant. Now simplifying this we get cos theta into cos phi into cos phi minus this product but already minus is present now this minus is converted to plus sin phi cos theta sin phi minus next term is minus sin theta. So it is minus minus plus sin theta deleting the third row second column we get this determinant now evaluating it we get sin theta cos phi into cos phi minus but already minus is present so it is plus sin theta sin phi into sin phi and the last term is 0 therefore the value of the third term is 0. Now it is equal to r square into sin theta. Now here we can observe that it is cos phi into cos phi is cos square phi and here it is sin phi into sin phi is sin square phi from these two terms cos theta is a common term we can take it outside that term converted to cos square theta and the remaining quantity in the bracket is cos square phi plus sin square phi was value is 1. So simply we get the term as cos square theta plus again in the second bracket also we can see that after taking sin theta common from these two terms the remaining quantity in the bracket is cos square phi plus sin square phi was value is 1 as we have taken sin theta common already outside sin theta is present therefore sin theta into sin theta it is sin square theta now again the value of cos square theta plus sin square theta is 1 therefore 1 into r square sin theta is r square sin theta.