Question 1003757


(a)
{{{f(x)=x^2+x}}}

{{{f}}}'{{{(x) = 2x+1}}}

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


using the {{{(f(x+h)-f(x))/h }}}method:

{{{lim( h->0)}}}{{{(f(x+h)-f(x))/h  }}}


={{{lim h->0}}}{{{(1/(x+h-1)-1/(x-1))/h }}}


={{{lim h->0}}}{{{(1(x-1)/((x+h-1)(x-1)))-1(x+h-1)/((x-1)(x+h-1)) )/h}}}


={{{lim h->0}}}{{{((x-1)-(x+h-1))/(h(x-1)(x+h-1))}}}


={{{lim h->0}}}{{{-h/(h(x-1)(x+h-1))}}}


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


so,
{{{f}}}'{{{(x) = -1/(x-1)^2}}}