How do I us the Limit definition of derivative on #f(x)= 1/x#?

Question

4947 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-10-03T04:18:08

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

#f(x)=1/x#

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

Make the substitutions for #f(x)# and #f(x+h)#

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

Find the least common denominator

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

Simplify the numerator of the complex fraction

#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#

Distribute the negative in the numerator

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

Simplify the numerator

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

Division is equivalent to multiplying by the reciprocal

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

Cancel the factors of #h# and simplify

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

Substitute in the value of 0 for #h# and simplify

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

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

#=-1/x^2#

See the video below.