How do you find f'(x) using the limit definition given # (1/x^2) #?

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Answer 1 · 2016-07-05T11:34:52

#= - 2/x^3 #

#f(x) = 1/x^2#

#f'(x) = lim_{h to 0} (f(h+h) - f(x))/(h)#

#= lim_{h to 0} 1/h * (1/(x+h)^2 - 1/x^2)#

combining fractions:

#= lim_{h to 0} 1/h * (x^2- (x+h)^2)/(x^2(x+h)^2) #

#= lim_{h to 0} 1/h * (x^2- (x^2+2hx + h^2))/(x^2(x+h)^2 #

#= lim_{h to 0} 1/h * (-2hx - h^2)/(x^2(x+h)^2 #

#= lim_{h to 0} (-2x - h)/(x^2(x+h)^2 #

#= lim_{h to 0} (-2x )/(x^2(x+h)^2) + mathcal(O)(h)#

#= lim_{h to 0} (-2 )/(x(x+h)^2) + mathcal(O)(h)#

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

#= - 2/x^3 #