Question 1010141
{{{f(x) = 1/(x-1)}}}



----------------------------



Use f(x) to find f(x+h)



{{{f(x) = 1/(x-1)}}}



{{{f(x+h) = 1/(x+h-1)}}}



----------------------------



Compute the difference quotient



{{{(f(x+h)-f(x))/h = (1/(x+h-1)-1/(x-1))/h}}}



{{{(f(x+h)-f(x))/h = ((x-1)/((x-1)(x+h-1))-(x+h-1)/((x-1)(x+h-1)))/h}}}



{{{(f(x+h)-f(x))/h = (((x-1)-(x+h-1))/((x-1)(x+h-1)))/h}}}



{{{(f(x+h)-f(x))/h = (((x-1)-(x+h-1))/((x-1)(x+h-1)))*(1/h)}}}



{{{(f(x+h)-f(x))/h = ((x-1-x-h+1)/((x-1)(x+h-1)))*(1/h)}}}



{{{(f(x+h)-f(x))/h = (-h)/(h(x-1)(x+h-1))}}}



{{{(f(x+h)-f(x))/h = (-cross(h))/(cross(h)(x-1)(x+h-1))}}}



{{{(f(x+h)-f(x))/h = (-1)/((x-1)(x+h-1))}}}



----------------------------



The difference quotient is {{{(f(x+h)-f(x))/h = (-1)/((x-1)(x+h-1))}}}



Now evaluate the limit as h goes to 0. Since we no longer have to worry about division by zero errors, this is the equivalent to plugging in h = 0 to get 



{{{(-1)/((x-1)(x+h-1))=(-1)/((x-1)(x+0-1))=(-1)/((x-1)(x-1)) = (-1)/((x-1)^2)}}}



----------------------------



So the derivative of f(x) is f ' (x) = {{{ (-1)/((x-1)^2)}}}