Question
Find a formula for the derivatives of the function m(x)=1/(x+1)
IS the answer -1/(x+h)(x+h+1)
IS the answer -1/(x+h)(x+h+1)
Answers
Damon
[(x+1)(0) - 1(1)]/(x+1)^2
=-1/(x+1)^2
=-1/(x^2+2x+1)
now doing it as a limit you must let h go to zero
[f(x+h) -f(x)]/h as h -->0
[1/(x+h+1)-1/(x+1)]/h
[(x+1)-(x+h+1)]/[h(x+h+1)(x+1)]
-1/(x+h+1)(x+1)
as h -->0
-1/[(x+1)(x+1)]
-1/(x^2+2x+1) as we knew
=-1/(x+1)^2
=-1/(x^2+2x+1)
now doing it as a limit you must let h go to zero
[f(x+h) -f(x)]/h as h -->0
[1/(x+h+1)-1/(x+1)]/h
[(x+1)-(x+h+1)]/[h(x+h+1)(x+1)]
-1/(x+h+1)(x+1)
as h -->0
-1/[(x+1)(x+1)]
-1/(x^2+2x+1) as we knew