How do you find the formula for the derivative of #1/x#?

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Answer 1 · 2015-04-14T18:07:15

#(d x^a)/(dx) = a*x^(a-1)#

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

Therefore
#(d 1/x)/(dx) = -x^(-2)# or #- 1/(x^2)#

Answer 2 · 2015-04-14T18:14:14

I will assume that you are working from first principles, that is the definition.

There are two choices for how to "officially" define the derivative, every textbook author and teacher makes a decision which one to make official.

I'll use: #f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h#

For #f(x) = 1/x#, we get:

#f'(x) = lim_(h rarr0) (f(x+h)-f(x))/h#

#color(white)"ssssss"##= lim_(h rarr0) (1/(x+h)-1/x)/h#

#color(white)"ssssss"##= lim_(h rarr0) ((x-(x+h))/(x(x+h)))/(h/1)#

#color(white)"ssssss"##= lim_(h rarr0) ((x-x-h))/(x(x+h))*(1/h)#

#color(white)"ssssss"##= lim_(h rarr0) ((-1))/(x(x+h))#

#color(white)"ssssss"##= -1/x^2#

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