 Hello and welcome to the session, let's work out the following problem. It says using differentials find the approximate value of the following up to three places of decimal and the given number is under the root 0.6. So let's now move on to the solution and let us first express y as a function of x. And here we choose x in such a way so that we can easily find out the square root. So here we choose x plus delta x as the number itself is 0.6 and we choose x as 1. Therefore delta x would be minus of 0.4. Now we know that delta y is equal to f of x plus delta x minus fx. Now f of x plus delta x is 0.6 minus fx that is under the root 1. Now we have delta y is equal to under the root 0.6 minus 1. So this implies under the root 0.6 is equal to delta y plus 1. And here we write that delta y is equal to under the root x plus delta x minus under the root x. Now delta y is approximately equal to dy and dy is equal to dy by dx into delta x. Now dy by dx is 1 by 2 into x to the power minus 1 by 2 into delta x. Delta x is minus 0.4. Now this is 1 by 2 into under the root x into minus 0.4. Now this is under the root 1 into minus 0.4. So this is equal to 1 by 2 into minus 0.4 and this is equal to minus 0.2. Now delta y is 0.2 minus 0.2. So under the root 0.6 is equal to delta y plus 1 that is minus 0.2 plus 1. So this is equal to 0.8. Hence square root of 0.6 is 0.8. So this completes the question and the session. Bye for now. Take care. Have a good day.