 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 here the number given to us is 26.57 to the power 1 by 3. So let's now start the solution and let us first define y as a function of x and here we define 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 and that is near to 26.57 so we choose x to be 27 and since we need to have x plus delta x as 26.57 we choose delta x as minus of 0.43 so x plus delta x is equal to 26.57 now we know that delta y is f of x plus delta x minus f x so this is x plus delta x to the power 1 by 3 minus x to the power 1 by 3 so this is 26.57 to the power 1 by 3 minus 27 to the power 1 by 3 so this implies 26.57 to the power 1 by 3 equal to delta y plus 27 to the power 1 by 3 now 27 can be written as 3 cube as power 1 by 3 so we have 26.57 to the power 1 by 3 equal to delta y plus 3 now we have to find the value of delta y and we know that delta y is approximately equal to dy dy is equal to dy by dx into delta x so delta y is equal to dy by dx into delta x 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 minus 2 by 3 into delta x and delta x is minus 0.43 now this is again 1 by 3 into x to the power minus 2 by 3 now x is 27 27 is 3 cube this is equal to 1 by 3 into 3 to the power 2 and 3 to the power 2 is 9 so this is 3 into 9 into 0. minus 0.43 this is equal to minus 0.43 upon 27 which is equal to minus 0.159 now 26.57 to the power 1 by 3 is equal to delta y plus 3 and delta y is this this is minus 0.159 plus 3 and this is equal to 2.984 hence the approximate value of 26.57 to the power 1 by 3 is 2.984 so this completes the question and the session by for now take care have a good day