 Hello and welcome to the session. My name is Mansi and I am going to help you with the following question. The question says find the derivative of the following function from first principle 1 by x square. Let us start with the solution to this question. We know that according to first principle f dash x is equal to limit h approaching to 0 function at x plus h minus function at x divided by h we have fx is equal to 1 by x square. So function at x plus h will be equal to 1 upon x plus h the whole square. Now we have to find f dash x that will be equal to limit h approaching to 0 function at x plus h is 1 upon x plus h the whole square minus function at x is 1 by x square this divided by h. Now this is equal to limit h approaching to 0. Now in the denominator of the numerator we will have x square into x plus h the whole square and here we will have x square minus x plus h the whole square the whole divided by h this can be written as limit h approaching to 0 x square minus x square minus h square minus 2xh divided by x square into x plus h the whole square into h. Now plus x square gets cancelled with minus x square and we have limit h approaching to 0 from these two terms we can take minus h common so we have minus h into h plus 2x divided by x square into x plus h the whole square into h. Now this h gets cancelled with this h and we have limit h approaching to 0 minus of h plus 2x divided by x square into x plus h the whole square. Now we apply the limit we put h equal to 0 in the numerator and denominator we will have minus 2x divided by x square into x plus 0 the whole square is x square so this will be equal to minus 2 into x divided by x cube into x we can write this as this also x gets cancelled with x and we have minus 2 by x cube or we can simply write minus 2 by x cube so this is our answer to the question. I hope that you understood the question and enjoyed the session have a good day.