 Hello and welcome to the session. Let's work out the following question. It says using differential find the approximate value of each of the following up to three places of decimals. Before moving on to the solution, let us first understand the theory of approximation symmetrically. Suppose we have a function y is equal to fx and we have any point on the curve say pxy. Now since y is fx, so the point is pxfx. Suppose we give a small increment to x, so the point shifts somewhere here. Now the coordinates of this point would be x plus delta x, the increment is delta x, so the x coordinate will be x plus delta x and the y coordinate will be y plus delta y. Now since y is function of x, so delta y is the increment in y corresponding to the increment in x, so delta y is equal to f of x plus delta x minus fx. Now clearly this distance is delta x and this is delta y. So delta x is same as dx that is differential of x and this distance is dy that is differential of y and we assume that delta y is approximately equal to dy. Let us now move on to the solution. Now first of all we need to define y as a function of x, so we define y as a function of x. Since here we have to find the square root of 25.3 we define it as y is equal to under the root x where x is 25 and we choose this x in such a way so that we can easily find out its square root. Now we know that delta y is f of x plus delta x minus fx, so we choose delta x to be 0.3 so that x plus delta x is 25.3, so x plus delta x is equal to 25.3. Now delta y is equal to f of x plus delta x minus fx, so delta y is equal to f of x plus delta x that is under the root of 25 plus 0.3 that is 25.3 minus fx that is under the root of x and x is 25 so it is 25. Now we have to find the value of under the root 25.3, so we have under the root 25.3 equal to delta y plus the square root of 25 is 5. Now we need to find the value of delta y but we know that delta y is approximately equal to dy, so dy is equal to dy by dx into delta x since delta x is equal to dx. Now y is under the root x so dy by dx would be 1 by 2 into x to the power minus 1 by 2 into delta x, delta x is 0.3 that simplifies dy is equal to 1 by 2 into under the root x into 0.3, this simplifies dy is equal to 1 by 2 into root x, x is 25 into 0.3 square root of 25 is 5, so 5 into 2 is 10 into 0.3 so this is equal to 0.03. Now under the root 25.3 is equal to delta y plus 5, so under the root 25.3 is equal to delta y plus 5 or delta y is 0.03 plus 5 so this is equal to 5.03. Hence the approximate value of under the root 25.3 is 5.03. So this completes the question and the session. Bye for now. Take care. Have a good day.