 Hello friends, let's work out the following problem. It says using differentials find the approximate value of the following up to three places of decimal so let's now move on to the solution and Let us first define y as a function of x So we define y as x to the power 1 by 2 and here we choose x in such a way So that we can easily find out its square root and that is also near to 0.0037 so we choose it to be 0.0036 and since we need to have x plus delta x as 0.0037 we choose delta x as 0.0000 1 right So x plus delta x is equal to 0.0037 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 2 minus x to the power 1 by 2 So this is equal to 0.0037 to the power 1 by 2 minus 0.0036 To the power 1 by 2 so this implies 0.0037 to the power 1 by 2 is equal to delta y Plus 0.0036 to the power 1 by 2 Now 0.0036 can be written as 0.06 square It's power 1 by 2 2 to get cancelled and this is equal to delta y Plus 0.06 Now we know that Delta y is approximately equal to D y and D y is D y y dx into delta x. So delta y is equal to D y y dx into delta x Now a y is x to the power 1 by 2 So D y y dx is equal to 1 by 2 into x to the power minus 1 by 2 into delta x and delta x is 0.00 0 1 Now this is equal to 1 by 2 into x to the power 1 by 2 into 0.000 1 let's now substitute the value of x here. So we have 1 by 2 into x is 0.0036 and 00.36 can be written as 0.006 square It's power 1 by 2 into 0.000 1 This is equal to 1 upon 2 into 0.06 which is equal to 0.12 Into 0.0001 and this is equal to 0.00 083 so delta y is this now we know that 0.0037 To the power 1 by 2 is equal to Delta y plus 0.06 and delta y is this it is delta y plus 0.06 This is equal to 0.060 8 3 Approximately hence the value of 0.0037 to the power 1 by 2 is 0. 060 So this completes the question and the session bye for now take care have a good day