00:01
Hello everyone in this problem we have given a function that is f of x is equals to 1 by x plus 1 so here its first order derivative that is f dash x will become minus of 1 by x plus 1 raise to par 2 in the similar way the second order derivative that is f double dash x will become 2 by x plus 1 raise to par 3 so it can also be rearranged as minus 1 raise to power 2 into 2 divided by x plus 1 raised to power 2 plus 1.
00:36
In the similar way, the next order derivative, that is third order, will become minus 6 divided by x plus 1 raised 2 power 4.
00:47
So here it can also be rearranged minus 1 raised 2 par 3 into factorial 3 divided by x plus 1 raised to part 3 plus 1.
00:58
So here if we generalize the fourth order derivative will also become minus 1 raised to par 4 into factorial 4 divided by x plus 1 raised to par 4 plus 1.
01:14
In the similar way, the nth order derivative that is f of n x will become minus 1 raised to par n into n factorial divided by x plus 1 raised to par n plus 1.
01:31
So here, as we know by methodorial theorem, the remainder that is r and x that is equals to f of n plus 1 of z divided by n plus 1 factorial into x raised to par, n plus 1, where we can say the value of the z is less than x and greater than 0.
01:59
So here as we know for x that is equals to 0 .4, the error of approximation is 0 .01...