 Hello and welcome to the session the question says find the derivative of the following functions from the first principles. Second part is minus x raised to the power minus one. Let us now start with the solution and let us denote fx is equal to minus x raised to the power minus one which can further be written as one upon minus x. Now let us find the value of the function at the point x plus h to be one upon minus of x plus h. Now this can further be written as minus one upon x by multiplying the numerator and denominator by minus one and similarly the multiplying the denominator and numerator by minus one here we have minus one upon x plus h. Now let us find the first derivative of the given function which according to the first principle is given by limit as h approaches to 0 f at x plus h minus fx upon h. Now let us substitute the value of the function at the point x plus h and x. So we have minus one upon x plus h minus of minus one upon x upon h this is further equal to limit as h approaches to 0 minus one upon x plus h plus one upon x upon h. This further equal to limit as h approaches to 0 for solving the numerator here we have x into x plus h minus of x plus x plus h upon h. This further equal to limit as h approaches to 0 minus x plus x on cancelling we have h upon x into x plus h into one upon h given further simplifying we have limit as h approaches to 0 one upon x square plus x h. This is further equal to one upon x square since with h approaches to 0 this terms become 0 and we are left with one upon x square. Thus our answer is with the help of first principle derivative of minus x face to the power minus one as one upon x square. So this completes the second part Hope you have understood it. Take care and have a good day.