 Hello everyone myself, Mrs. Mayuri Kangal, Assistant Professor of Mathematics from the Department of Humanities and Sciences, Valjan 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 previous video we have seen the partial differentiation of first order. Now in this video we will see the partial differentiation of higher order. The partial derivatives of higher order if they exist can be obtained from the partial derivatives of first order by differentiating them again. Since first order partial derivatives dou z by dou x and dou z by dou y themselves are the functions of x and y they can be further differentiated partially with respect to x as well as y. These are called the second order partial derivatives of z. Now let us see the notations for the second order partial derivatives. Let f of x y be given function such that dou f by dou x and dou f by dou y that is the first order partial derivatives exist. Then differentiating dou f by dou x with respect to x partially that is dou by dou x of dou f by dou x is dou 2f by dou x square or f to the suffix is xx. Similarly dou f by dou y when differentiated partially with respect to x we get dou 2f by dou x dou y which is also denoted as f to the suffix xy. Now if dou f by dou x is differentiated partially with respect to y we get dou 2f by dou y dou x that is f to the suffix yx and dou f by dou y when differentiated partially with respect to y we get dou 2f by dou y square f to the suffix is yy. In general f of x y be a function of two independent variables x and y. When the function f of x y is differentiated partially with respect to x we get dou f by dou x and when this function is differentiated partially with respect to y we get dou f by dou y these two derivatives are called as the first order partial derivatives. Now when dou f by dou x is differentiated partially with respect to x we get dou 2f by dou x square and when dou f by dou x is differentiated partially with respect to y we get dou 2f by dou y dou x. Similarly when dou f by dou y is differentiated partially with respect to x we get dou 2f by dou x dou y and when dou f by dou y is differentiated partially with respect to y, we get dou 2f by dou y square. These four derivatives are called as the second order partial derivatives and out of these four second order partial derivatives, the partial derivatives dou 2f by dou y dou x and dou 2f by dou x dou y are called as the mixed partial derivatives. This point is written in this note, note that in general these two mixed partial derivatives are same. Now we will see the order of differentiation. In all the cases of higher order differentiation, we differentiate moving along the denominator from right to left. See here, f to the suffixes xx or dou 2f by dou x square, we have to move from right that is here the first differentiation will be with respect to x then again with respect to x. Similarly, when we have to find out f to the suffix xy or dou 2f by dou x dou y, we have to again move from right to left. In the right, we have dou y, so the first differentiation will be with respect to y and then with respect to x. Similarly, for dou 2f by dou y dou x, the first differentiation will be with respect to x and then with respect to y and in dou 2f by dou y square, first differentiation is with respect to y and then again with respect to y. So, to decide the order of differentiation, we have to move along the denominator from right to left. Now, let us see the standard rules. If u and v are functions of x and y possessing partial derivatives of first order, then just like the ordinary differentiation, the rules of addition, subtraction, multiplication and division of ordinary derivatives holds good for the partial derivatives also. Now, pause the video for a minute and write the answer of this question. Find first order partial derivatives of f of xy equals to x square plus sin y cube. I hope you all have written the solution. Let us check the solution. Given that f of xy is equals to x square plus sin y cube, now differentiating this f partially with respect to x, treating y as constant gives us dou f by dou x is equals to the derivative of x square is 2x and as y is treated as constant, the derivative of this term will be 0. So, we get dou f by dou x equals to 2x. Now, we will differentiate f partially with respect to y gives us dou f by dou y. The derivative of x square will be 0 as x is treated as constant. So, we get dou f by dou y by differentiating sin y cube. So, we get dou f by dou y as 3y square cos y cube. Now, let us go for the examples. First example, find all the second order partial derivatives of f of xy equals to cos 2x minus x square e raised to 5y plus 3y square. First, we find the first order partial derivatives that is dou f by dou x and dou f by dou y. So, we will differentiate this f partially with respect to x as well as partially with respect to y. So, we get dou f by dou x equals to the derivative of cos 2x is minus sin 2x into 2. Now, e raised to 5y is treated as constant. So, e raised to 5y as it is and the derivative of x square is 2x. So, minus 2x e raised to 5y plus 3y square is there. So, it is treated as constant and its derivative is 0. So, we get dou f by dou x as minus 2 sin 2x minus 2x e raised to 5y. We will call it as equation number 1. Now, dou f by dou y is obtained by differentiating partially with respect to y treating x as constant. So, the derivative of cos 2x will be 0. Now, minus x square e raised to 5y is there and the x square is treated as constant. So, minus x square as it is and the derivative of e raised to 5y is e raised to 5y into 5 plus the derivative of 3y square is 6y. So, we get dou f by dou y as minus 5x square e raised to 5y plus 6y. We will call it as equation number 2. Now, differentiating equation 1 with respect to x partially treating y as constant gives us dou 2f by dou x square. So, dou by dou x of dou f by dou x. Now, we will substitute dou f by dou x using equation 1. So, we will write it as dou by dou x of minus 2 sin 2x minus 2x e raised to 5y. The differentiation is with respect to x treating y as constant. So, f xx is equals to minus 2 as it is the derivative of sin 2x is cos 2x into 2 minus sin as it is e raised to 5y into 2 is a constant. So, 2 e raised to 5y as it is and the derivative of x is 1. So, we get f xx as minus 4 cos 2x minus 2 e raised to 5y. Similarly, differentiating equation number 2 with respect to y partially treating x as constant it gives us dou 2f by dou y square. So, to obtain dou 2f by dou y square we will differentiate dou f by dou y partially with respect to y. So, dou by dou y of minus 5x square e raised to 5y plus 6y. The differentiation is with respect to y treating x as constant. So, for the first term minus 5x square will be treated as constant. So, in f yy minus 5x square as it is and the derivative of e raised to 5y is e raised to 5y into 5 plus the derivative of 6y is 6. So, we get f yy as minus 25x square e raised to 5y plus 6. Similarly, differentiating equation 1 with respect to y partially treating x as constant which gives us dou 2f by dou y dou x. That is dou f by dou x is differentiated partially with respect to y. Now, for this we will substitute dou f by dou x. So, we get dou by dou y of minus 2 sin 2x minus 2x e raised to 5y. The differentiation is with respect to y treating x as constant. So, minus 2 sin 2x will be a constant term and its derivative will be 0. For the second term minus 2x will be treated as constant. So, it will be kept as it is and the derivative of e raised to 5y is e raised to 5y into 5 which gives us f yx equals to minus tan x e raised to 5y. And differentiating equation 2 with respect to x partially treating y as constant which gives us dou 2f by dou x dou y. That is dou f by dou y is differentiated partially with respect to x. So, substituting dou f by dou y gives us the equation dou by dou x of minus 5x square e raised to 5y plus 6y. Now, the differentiation is with respect to x treating y as constant. So, minus 5 e raised to 5y will be constant and the derivative of x square will be 2x. So, we get f xy equals to minus tan x e raised to 5y and the derivative of 6y will be 0 as y is treated as constant. So, we get f xy equals to minus tan x e raised to 5y. If you observe this f yx is minus tan x e raised to 5y and f xy is also minus tan x e raised to 5y which illustrate that the mixed partial derivatives are same. Thank you.