 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 3 places of the decimal. So let's now proceed on to the solution and let us first define y as a function of x and here we take y equal to x to the power 1 by 3 and here we choose x in such a way so that we can easily find out its cube root. So we take it to be 27 and since we need to have x plus delta x as 26. So we will take delta x as minus 1 so x plus delta x is equal to 26. Now we know that delta y is equal to f of x plus delta x minus fx so this is x plus delta x to the power 1 by 3 minus x to the power 1 by 3. Now this is 26 to the power 1 by 3 minus 27 to the power 1 by 3 so this implies 26 to the power 1 by 3 is equal to delta y plus 27 to the power 1 by 3 now 27 can be written as 3 cube its power is 1 by 3 3 gets cancelled with 3 and we have 26 to the power 1 by 3 equal to delta y plus 3. Now we know that delta y is approximately equal to dy and dy is equal to dy by dx into delta x so delta y is equal to dy by dx into delta x and delta x is minus 1. Now y is x to the power 1 by 3 so dy by dx is equal to 1 by 3 into x to the power 1 by 3 minus 1 that is minus 2 by 3 into delta x which is minus 1 so this is equal to 1 by 3 into x to the power 2 by 3 into minus 1 now substitute the value of x so we have 3 into 27 to the power 2 by 3 27 can be written as 3 to the power 3 and its power is 2 by 3 into minus 1 now 3 gets cancelled with 3 and we have 1 by 3 into 3 square that is 9 into 3 that is 27 and into minus 1 it gives us minus 1 by 27 so delta y is minus 1 by 27 now 26 to the power 1 by 3 is equal to delta y plus 3 delta y is minus 1 upon 27 so this is minus 1 upon 27 plus 3 and it is equal to 80 by 27 and it is equal to 2.9629 and the approximate value of 26 to the power 1 by 3 is 0 2.9629 so that's all for this session goodbye and take care.