 Hello everyone, myself as Falmari. In the last two videos, we have discussed the definition of Jacobian and one property of the Jacobian. In this video, we will discuss one more property of the Jacobian that is called the Jacobian of composite functions. Learning outcome, at the end of this session, students will be able to find the Jacobian of composite functions. This is the second property. If u, v are functions of two independent variables r, s and r, s are functions of two new independent variables x, y, that is it directly indicating that u, v are composite functions of x, y. Therefore, the Jacobian of u, v with respect to x, y is equal to Jacobian of u, v with respect to r, s into Jacobian of r, s with respect to x, y. Pause this video and answer this question. If u, v, w are functions of x, y, z and x, y, z are functions of u, v, w then what is the value of the product of these two Jacobians? I hope that all of you have written the answer. Now by the property number one, the value of the j into j dash equal to 1. Let us consider some examples. Example number one, if x equal to a into the bracket u plus v, y equal to b into the bracket u minus v and u equal to r square cos theta, v equal to r square sin theta then find Jacobian of x, y with respect to r and theta. Now, here we can see that x and y are composite functions of r and theta. Therefore, the Jacobian of x, y with respect to r theta is provided by the product of the Jacobian of x, y with respect to u, v and Jacobian of u, v with respect to r, theta. Let us find out one by one. Now, first we will find out Jacobian of x, y with respect to u, v which is defined by the 2 by 2 determinant and it is equal to now differentiating x partially with respect to u treating v constant. Now, here is x equal to a into u plus v, a is constant derivative of u is 1 and derivative of v is 0. Therefore, a into 1 is simple a and differentiating x partially with respect to v, a as it is into derivative of u with respect to v is 0 and the derivative of v is 1 and 1 into a is a. Now, the next entry is partial derivative of y with respect to u. Now, here y is b into bracket u minus v, v is constant, derivative of u is 1 and derivative of v with respect to u is 0, therefore, it is b and differentiating y partially with respect to v, we get it as minus b. Now, simplifying this determinant, a into minus b is minus a b minus b into a that is minus a b which is equal to minus of 2 a b. Now, we will find out this next determinant, next Jacobian. Here, we have given u equal to r square cos theta v equal to r square sin theta. Now, Jacobian of u v with respect to r theta is defined by this 2 by 2 determinant. Now, let us find out all these partial derivatives. Differentiating u partially with respect to r treating theta constant as theta constant we can write down cos theta as it is and the derivative of r square is 2 r. Now, differentiating u partially with respect to theta treating r constant, as r constant we can write r square as it is and the derivative of cos theta is minus sin theta. Now, differentiating v partially with respect to r treating theta constant, we write sin theta as it is and the derivative of r square is 2 r. And, finally differentiating v partially with respect to theta treating r constant as r constant r square as it is and the derivative of sin theta is cos theta. Now evaluating now this determinant 2r cos theta into r square cos theta is 2r cube cos square theta minus but already minus is present in this product. So, we can replace it by plus 2r sin theta into r square sin theta will be 2r cube sin square theta. Now we can take 2r cube common from these two terms and the quantity left in the bracket is cos square theta plus sin square theta and it has the value 1 therefore the value of the determinant is 2r cube. Now the Jacobian of x comma y with respect to r comma theta is given as Jacobian of x comma y with respect to u comma v into Jacobian of u v with respect to r comma theta is equal to the value of the first Jacobian we have obtained it as minus 2ab into and the value of the second Jacobian is 2r cube and finally we get the answer as minus 4ab r cube. Let us consider one more example if x equal to root with w y equal to root w z equal to root uv and u equal to r sin theta cos phi v equal to r sin theta sin phi w equal to r cos theta then find out Jacobian of x y z with respect to r theta phi. Now these relations are implying that x y z are composite functions of r theta phi therefore this Jacobian is provided by the product of the Jacobians of Jacobian of x y z with respect to uvw and Jacobian of uvw with respect to r theta phi we will find 1 by 1. Now initially we have provided x y z in terms of uvw now let us calculate the partial derivatives 1 by 1. Now first differentiating x partially with respect to u but u is not present in x therefore the derivative is 0. Now differentiating x partially with respect to v treating remaining variables constant now the derivative of the root we know that 1 upon 2 times root into derivative of vw with respect to v is w. We can write this w as root w into root w one root w is present in the denominator after removing that root w we get the derivative as root w upon 2 root v. Now differentiating x partially with respect to w treating u and v constant we get again it is also in terms of root therefore its derivative is 1 upon 2 times root v w into derivative of this v w with respect to w is v. Now we can remove root v from the numerator and denominator and we get the derivative as root v upon 2 root w. Now next differentiating y partially with respect to u treating v and w constant again it is in terms of the root function therefore its derivative is 1 upon 2 times root w u into derivative of w u with respect to u is w. After removing root w from the numerator and denominator we get it as root w upon 2 root u. Now y does not contains v therefore partial derivative of y with respect to v is 0. Now differentiating y partially with respect to w treating u and v constant we get it as u upon 2 root w u and after removing root u from the numerator and denominator we get it as root u upon 2 root w. Now finally differentiating z partially with respect to u treating v and w constant we get it as v upon 2 root u v and after removing root v from the numerator and denominator we get root v upon 2 root u. Now differentiating z partially with respect to v we get it as u upon 2 root u v. After removing root u from the numerator and denominator we get root u upon 2 root v and z will not contain w therefore partial derivative of z with respect to w is 0. Now by the definition Jacobian of XYZ with respect to u v w defined by this 3 by 3 determinant. Now substituting all the partial derivatives in the determinant we get this expression. Now evaluating this determinant along the first row the first entry is 0 therefore the value of the first term is 0 minus this term root w upon 2 root v into the bracket. Now deleting the first row and second column and simplifying the remaining determinant that is root w upon 2 root u into 0 is 0 minus root v upon 2 root u into root u upon 2 root w is minus root u v upon 4 root u w plus next term is root v upon 2 root w into deleting the first row last column and simplifying the remaining determinant we get it as root w upon 2 root u into root u upon 2 root v minus 0 into this quantity is 0. Now by multiplying by this minus root w upon 2 root v to this bracket we get the term as minus minus will be becomes plus root u v w upon 8 root u v w. Now again multiplying by this plus root v upon 2 root w to this bracket now it will be again root u v w upon 8 into root u v w. Now we can remove this root u v w from the numerator and denominator from both the terms we get it as 1 upon 8 plus 1 upon 8 and it is equal to 2 by 8 and 2 by 8 is nothing but 1 by 4. Now this is the value of this Jacobian. Now we will find out the next Jacobian we have provided u v w in terms of r theta and phi. Now differentiating u partially with respect to r treating remaining variables constant. Now the derivative of r is 1 and as the remaining variables are constant we can write down sin theta cos phi as it is. Differentiating u partially with respect to theta treating r and phi constant r phi constant we can write down r cos phi as it is. Differentiating u partially with respect to theta treating r and phi constant as r and phi are constant we can write down r cos phi as it is and the derivative of sin theta is cos theta and finally differentiating u partially with respect to phi treating r theta constant as r theta is constant we can write down r sin theta as it is and the derivative of cos phi is minus sin phi. Now taking the partial derivative of v with respect to r now the derivative of r is 1 and sin theta of sin phi are constant, we can write them as it is. Now, taking the partial derivative of v with respect to theta treating r and phi constant, as r phi constant we can write down r sin phi as it is and the derivative of sin theta is cos theta. And finally, differentiating v partial with respect to phi treating r and theta constant r theta constant therefore, we can write down r sin theta as it is and the derivative of sin phi is cos phi. Last function differentiating w with respect to r treating theta and phi constant, theta constant we can write down cos theta as it is and the derivative of r is 1. And differentiating w with respect to theta treating r and phi constant as r is constant we can write down r as it is and the derivative of this cos theta is minus sin theta and w is not containing phi therefore, dou w by dou phi is 0. Now, the definition of Jacobian for u v w with respect to r theta phi is this 3 cross 3 determinant, substituting all these partial derivatives here we get this big determinant. Now, from the second column and third column we can observe that from the second column it is possible to take r common and from the last column it is possible to take r sin theta common. After taking this common this determinant is reduces to r square sin theta into this big determinant. Now, simplifying this determinant along the last row we get it as r square sin theta. Now, the finally, the value of Jacobian of x y z with respect to r theta phi is provided by the product of these Jacobians. Now, the value of the first Jacobian is 1 by 2 and the value of the second Jacobian is r square sin theta. Now, it is the required answer.