How do you find f'(x) using the definition of a derivative #f(x) =1/(x-3)#?

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Answer 1 · 2018-03-20T03:23:15

#f'(x) = -(1)/(x-3)^2#

We are given:

#f(x) = 1/(x-3)#

The limit definition of the derivative is:

#f^(')(x) = lim_(\Deltax->0) (f(x+\Deltax)-f(x))/(\Deltax)#

#=lim_(\Deltax->0)(1/(x+\Deltax-3)-1/(x-3))/(\Deltax)#

#=lim_(\Deltax->0)((x-3)/((x+\Deltax-3)(x-3))-(x+\Deltax-3)/((x+\Deltax-3)(x-3)))/(\Deltax)#

#=lim_(\Deltax->0)((x-3-(x+\Deltax-3))/((x+\Deltax-3)(x-3)))/(\Deltax)#

#=lim_(\Deltax->0)(cancelxcancel(-3)cancel(-x)-\Deltaxcancel(+3))/(\Deltax(x+\Deltax-3)(x-3))#

#=lim_(\Deltax->0)-(cancel(\Deltax))/(cancel(\Deltax)(x+\Deltax-3)(x-3))#

#=lim_(\Deltax->0)-(1)/((x+\Deltax-3)(x-3))#

#=-(1)/((x+(0)-3)(x-3))#

#=-(1)/((x-3)(x-3))#

#=-(1)/(x-3)^2#

Hence:

#=> color(green)(f'(x) = -(1)/(x-3)^2)#