 Hello everyone. Myself A.S. Falmari, assistant professor, Department of Humanities and Sciences, Balchan Institute of Technology, Solapur. In the last video, we have discussed the definition and some examples of the Jacobian. Now in this video, we consider some properties of the Jacobian and the examples based on that properties. The learning outcome of this session is at the end of this session, students will be able to find the Jacobian of given functions. Now let us consider the properties of the Jacobian. Now the first property is if j is the Jacobian of u v with respect to x y and j dash is the Jacobian of x y with respect to u v, then j into j dash is equal to 1. That is, Jacobian of u v with respect to x y into Jacobian of x y with respect to u v is equal to 1. Now pause this video and write down the formula for finding the Jacobian of u v w with respect to x y z. I hope that all of you have written the answer. The Jacobian of u v w with respect to x y z is defined by the 3 x 3 determinant in which the first row contains the partial derivatives of u with respect to x y z. Second row contains the partial derivatives of v with respect to x y z and finally the third row contains the partial derivatives of w with respect to x y z. Let us consider examples. Example number 1 if x equal to u into cos v y equal to u into sin v prove that Jacobian of u v with respect to x y into Jacobian of x y with respect to u v is equal to 1. Solution Here we have given x and y in terms of u v. Therefore, first we find the Jacobian of x y with respect to u and v. Now, differentiating x partially with respect to u treating v constant we get now cos v is constant and the derivative of u with respect to u is 1. Now, differentiating x partially with respect to v treating u constant as u constant we can write u constant as it is and the derivative of cos v is minus sin v. Now, consider the next function y differentiating y partially with respect to u treating v constant as v constant we can write sin v as it is into the derivative of u with respect to u is 1. And finally, partial derivative of y with respect to v treating u constant is equal to as u is constant we can write u as it is and the derivative of sin v is cos v. Now, by the definition Jacobian of x y with respect to u v is defined by this 2 by 2 determinant. Now, substituting all these partial derivatives in this determinant we get this determinant. Now, evaluating it cos v into the bracket u into cos v minus sin v into into the bracket minus u into sin v. Now, after multiplying by cos v to this bracket we get it as u into cos square v and this minus minus will be plus u into sin v into sin v is sin square v. Now, let us take u common from these 2 terms we get it as u into the bracket cos square v plus sin square v and this bracket has the value 1. Therefore, 1 into u is u. Therefore, this u is the value of Jacobian of x y with respect to u v. Now, our next time is to find out the Jacobian of u v with respect to x y. In order to find this Jacobian we must have u v as a function of 2 independent variables x and y. In order to obtain this we have to solve these 2 equations for u and v. Now, first let us consider the ratio y by x which is equal to where y is u sin v and x is u cos v. Here we can remove this u and sin v upon cos v is tan v. In order to have a expression in terms of v, let us operate tan inverse on both sides. So, we get it as v equal to tan inverse of y by x. Now, here we have obtained v as a function of x y. Now, let us differentiate this v partially with respect to x treating y constant. Now, here v is tan inverse of y by x. Now, it has the derivative 1 upon 1 plus y by x bracket square that is y square upon x square into the derivative of y by x with respect to x is minus y upon x square. Now, after cross multiplication we get this term as x square upon x square plus y square into now this minus we can write here minus y by x square. Now, we can remove this x square and finally we get the derivative as minus y upon x square plus y square. Again, differentiating v partially with respect to y treating x constant. Now, the derivative is 1 upon 1 plus y by x square that is y square upon x square into the derivative of y by x with respect to y is 1 upon x. Now, after cross multiplication we get this term as x square upon x square plus y square into the bracket 1 upon x. Now, here we can remove 1x from the numerator and denominator and we get the derivative as x upon x square plus y square. Now, in the next one now we have to express u in terms of x y to express this now consider the relation x square plus y square we know that x is u cos v therefore, x square is u square cos square v plus and we know that y is u sin v therefore y square is u square sin square v. From these two terms we can take u square common and then we get u square into the bracket cos square v plus sin square v. Now, we know that cos square v plus sin square v has the value 1 therefore, we get the final relation as u square is equal to x square plus y square. Now, differentiating this relation partially with respect to x treating y constant we get 2u into dou u by dou x is equal to now the derivative with respect to x is 2x now dividing this relation by 2u we get dou u by dou x is equal to x upon u. Now, differentiating this relation partially with respect to y treating x constant we get 2u into dou u by dou y is equal to derivative of x square plus y square with respect to y treating x constant is 2y. Dividing this relation by 2u we get dou u by dou y is equal to y by u. Now, we have obtained all these four derivatives by the definition Jacobian of u v with respect to x y is defined by these 2 by 2 determinant Now, substituting all these four partial derivatives in this determinant we get this determinant Now, simplifying it we get x by u into the bracket x upon x square plus y square minus y by u into the bracket minus y upon x square plus y square. Now, x upon u into this bracket we get it as x square upon u into the bracket x square plus y square. minus minus is plus y into y is y square upon u into the bracket x square plus y square. Now, here the denominators are same therefore, we can add both these quantities we get it as the numerators are x square plus y square divided by u into the bracket x square plus y square. Now, we can remove this x square plus y square and we get the final Jacobian as 1 upon u. Now, let us consider the required product Jacobian of x y with respect to u v into Jacobian of u v with respect to x y is equal to the first Jacobian has the value u into and the second Jacobian has the value and the second Jacobian has the value 1 upon u and which is equal to 1 hence proved.