 Hello everyone, myself, Mrs. Mayuri Kangle, Assistant Professor of Mathematics from the Department of Humanities and Sciences, Vyalchand 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. Before going to start the examples, please pause the video for a minute and solve the example. The question is, if r square equals to x square plus y square plus z square, then find dou r by dou x, dou r by dou y and dou r by dou z. I hope you all have solved the example. Let us check the solution. If r square equals to x square plus y square plus z square, it means that r is a function of three variables x, y and z. So we have to find out the three partial derivatives, dou r by dou x, dou r by dou y and dou r by dou z. To find the dou r by dou x, we will differentiate this equation partially with respect to x. The derivative of r square is 2r, which is multiplied by dou r by dou x equals to the derivative of x square is 2x. Here y and z are treated as constants. So we get 2r, dou r by dou x equals to 2x to get cancelled from both the sides which gives us dou r by dou x equals to x upon r. Similarly differentiating this equation partially with respect to y, we get 2r dou r by dou y equals to 2y. Here x and z are treated as constants and to get cancelled which gives us dou r by dou y equals to y upon r. Similarly differentiating this equation partially with respect to z, treating x and y as constant gives us 2r dou r by dou z equals to 2z which gives us dou r by dou z equals to z upon r. Now let us go for the examples. First example, if u equals to f of r where r square equals to x square plus y square then prove that dou 2u by dou x square plus dou 2u by dou y square equals to f double dash of r plus 1 upon r f dash of r. Now if you observe u is given as a function of single variable r and r is given as the function of two variables x and y. So here we have to find out du by dr and dou r by dou x dou r by dou y which gives us dou u by dou x equals to du by dr into dou r by dou x and dou u by dou y equals to du by dr into dou r by dou y. For this let us evaluate the values of du by dr, dou r by dou x and dou r by dou y. It is given that r square equals to x square plus y square. Differentiating this equation partially with respect to x, treating y as constant gives us 2r dou r by dou x equals to 2x which gives us dou r by dou x equals to x upon r. When we differentiate this equation partially with respect to y, treating x as constant we get 2r dou r by dou y equals to 2y from which we get dou r by dou y equals to y upon r. Let us call these equations as equation number one. Now u is given as a function of r so dou u by dou x is equals to the derivative of this f of r is f dash of r multiplied by dou r by dou x which gives us dou u by dou x equals to f dash of r into x upon r from equation number one. Now to get dou 2u by dou x square we will differentiate this dou u by dou x partially with respect to x which gives us dou 2u by dou x square equals to, now here f dash of r into x upon r is there so we will use the rule of u into v, we know that if z is equals to u into v then dou z by dou x is equals to u into dou v by dou x plus v into dou u by dou x. So dou 2u by dou x square is equals to f dash of r into the derivative of x upon r plus x upon r into the derivative of f dash of r that is f double dash of r into dou r by dou x. Now here we have to differentiate x upon r so we will use the rule if z is equals to u upon v then dou z by dou x is equals to v dou u by dou x minus u into dou v by dou x upon v square. Here when we differentiate x upon r we get x as u and r as v so r into derivative of x1 minus x into derivative of r that is dou r by dou x which is divided by r square. So we get dou 2u by dou x square equals to f dash of r into bracket r into 1 minus x into dou r by dou x upon r square plus x upon r f double dash of r into dou r by dou x. Now using equation 1 we will replace dou r by dou x by x upon r so we get the equation as dou 2u by dou x square equals to f dash of r upon r square into bracket r minus x into x upon r plus x upon r into f double dash of r into x upon r. Now we will evaluate this taking LCM here which gives us the denominator as r into r square so r cube. Now here in the numerator we get r square minus x square so here we will write as f dash of r upon r cube into bracket r square minus x square plus now here x upon r one more x upon r so it gives us x square upon r square into f double dash of r. We will call it as equation number 2. Since dou 2u by dou x square is equals to f dash of r upon r cube into bracket r square minus x square plus x square upon r square into f double dash of r. Similarly we can differentiate dou u by dou y partially with respect to y which gives us dou 2u by dou y square which is f dash of r upon r cube into bracket r square minus y square plus y square upon r square into f double dash of r. This equation is obtained just by replacing x by y here. When we simplify it we will get the same result. We will call it as equation number 3. Now we have to prove that dou 2u by dou x square plus dou 2u by dou y square equals to f double dash of r plus 1 upon r into f dash of r. For this we will consider the left side. Consider the LHS dou 2u by dou x square plus dou 2u by dou y square f dash of r upon r cube into bracket r square minus x square plus x square upon r square into f double dash of r plus f dash of r upon r cube into bracket r square minus y square plus y square upon r square into f double dash of r. Now we will simplify this. dou 2u by dou x square plus dou 2u by dou y square is equals to f dash of r upon r cube into bracket r square minus x square plus x square upon r square into f double dash of r plus f dash of r upon r cube into bracket r square minus y square plus y square upon r square into f double dash of r. If we see the first term and third term f dash of r upon r cube is common from the two terms. Taking common we get f dash of r upon r cube into bracket r square minus x square plus r square minus y square plus now if you observe the second and fourth term f double dash of r is common. So taking common f double dash of r we get in the bracket x square upon r square plus y square upon r square. So simplify it we get f dash of r upon r cube into bracket now r square plus r square is 2r square minus sign common into the bracket x square plus y square plus f double dash of r into bracket. Here the denominator is the same so we can write it as x square plus y square upon r square. As we know that x square plus y square is equals to r square we will replace this x square plus y square in both the terms by r square which gives us f dash of r upon r cube into bracket 2r square minus r square plus f double dash of r into r square upon r square. Now this r square get cancelled and here 2r square minus r square is single r square which gives us f dash of r upon r cube into r square plus f double dash of r. Here in the numerator it is r square and the denominator we have r cube so the r square from the denominator get cancelled with the numerator and we get dou 2u by dou x square plus dou 2u by dou y square equals to f dash of r upon r plus f double dash of r which is our required result. Thank you.