How do I use the definition of a derivative to find the derivative of #f(x)=1/x# at #x=3#?

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Answer 1 · 2014-10-01T03:03:08

#lim_(h->0) (f(x+h)-f(x))/h#

#f(x)=1/x#
#f(x+h)=1/(x+h)#

Substitute in these values

#lim_(h->0) (1/(x+h)-1/x)/h#

Get common denominator for the numerator of the complex fraction.

#f'(x)=lim_(h->0) (x/x*1/(x+h)-1/x*(x+h)/(x+h))/h#

#f'(x)=lim_(h->0) (x/(x(x+h))-(x+h)/(x(x+h)))/h#

#f'(x)=lim_(h->0) ((x-x-h)/(x(x+h)))/h#

#f'(x)=lim_(h->0) ((-h)/(x(x+h)))/h#

#f'(x)=lim_(h->0) (-h)/(x(x+h))*1/h#

#f'(x)=lim_(h->0) (-h)/(xh(x+h))#

#f'(x)=lim_(h->0) (-1)/(x(x+h))#

#f'(x)=(-1)/(x(x+0))#

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

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

#f'(3)=(-1)/((3)^2)=-1/9#

Alternative method

Now take the derivative of #f(x)# using the power rule.

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

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

#f'(3)=-1/(3)^2=-1/9#