Given:#f(x) = 1/(x - 1)#, then #f(x + h) = 1/(x + h - 1)#
Find f'(x)
#f'(x) = lim_(hto0){f(x + h) - f(x)}/h#
Substitute #1/(x + h - 1)# for #f(x + h)# and #1/(x - 1)# for #f(x)#:
#f'(x) = lim_(hto0){1/(x + h - 1) - 1/(x - 1)}/h#
Multiply by 1 in the form of #(x - 1)/(x - 1)#:
#f'(x) = lim_(hto0){1/(x + h - 1) - 1/(x - 1)}/h(x - 1)/(x - 1)#
Multiply numerators and denominators:
#f'(x) = lim_(hto0){(x - 1)/(x + h - 1) - (x - 1)/(x - 1)}/(h(x - 1))#
The second term in the numerator becomes 1:
#f'(x) = lim_(hto0){(x - 1)/(x + h - 1) - 1}/(h(x - 1))#
Multiply by 1 in the form of #(x + h - 1)/(x + h - 1)#:
#f'(x) = lim_(hto0){(x - 1)/(x + h - 1) - 1}/(h(x - 1))(x + h - 1)/(x + h - 1)#
Multiply numerators and denominators:
#f'(x) = lim_(hto0){((x - 1)(x + h - 1))/(x + h - 1) - (x + h - 1)}/(h(x - 1)(x + h - 1))#
I shall mark what cancels:
#f'(x) = lim_(hto0){((x - 1)cancel(x + h - 1))/cancel(x + h - 1) - (x + h - 1)}/(h(x - 1)(x + h - 1))#
Remove the canceled factors:
#f'(x) = lim_(hto0){(x - 1) - (x + h - 1)}/(h(x - 1)(x + h - 1))#
Distribute the -1 in the numerator:
#f'(x) = lim_(hto0){x - 1 - x - h + 1}/(h(x - 1)(x + h - 1))#
Combine like terms in the numerator:
#f'(x) = lim_(hto0){-h}/(h(x - 1)(x + h - 1))#
#-h/h# becomes -1:
#f'(x) = lim_(hto0){-1}/((x - 1)(x + h - 1))#
It is safe to let #hto0#:
#f'(x) = {-1}/((x - 1)(x - 1))#
Simplify:
#f'(x) = -1/(x - 1)^2#