How do you find the derivative of #f(x)=1/x^2# using the limit process?

1 Answer
Steve M
Nov 28, 2016

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

Explanation:

By definition of the derivative # f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h #
So with # f(x) = 1/x^2 # we have;

# f'(x)=lim_(h rarr 0) ( 1/(x+h)^2 -1/x^2 ) / h #
# :. f'(x)=lim_(h rarr 0)1/h*(x^2-(x+h)^2)/((x+h)^2x^2) #
# :. f'(x)=lim_(h rarr 0) (x^2-(x^2+2hx+h^2))/(h(x+h)^2x^2) #
# :. f'(x)=lim_(h rarr 0) (x^2-x^2-2hx-h^2)/(h(x+h)^2x^2) #
# :. f'(x)=lim_(h rarr 0) (-2x-h)/((x+h)^2x^2) #
# :. f'(x)=(-2x)/((x)^2x^2) #
# :. f'(x)=-2/x^3 #

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