By definition
`f'(x)=lim_(Delta x->0)(f(x+Delta x)-f(x))/(Delta x)`
So for `f(x)=1/x` we have
`f'(x)=lim_(Delta x->0)(1/(x+Delta x)-1/x)/(Delta x)=lim_(Delta X->0)((x-(x+Delta x))/((x+Delta x)x))/(Delta x)=`
`lim_(Delta x->0)(x-(x+Delta x))/((x+Delta x)x cdot Delta x)=lim_(Delta x->0)(-Delta x)/((x+Delta x)x cdot Delta x)=`
`=lim_(Delta x->0)-1/((x+Delta x)x)=-1/(x^2)`
So your solution is `f'(x)=-1/x^2`
This is a special case of `(x^n)'=nx^(x-1)` for `n=-1`.
See This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Already a member? Log in here.
By definition
`f'(x)=lim_(Delta x->0)(f(x+Delta x)-f(x))/(Delta x)`
So for `f(x)=1/x` we have
`f'(x)=lim_(Delta x->0)(1/(x+Delta x)-1/x)/(Delta x)=lim_(Delta X->0)((x-(x+Delta x))/((x+Delta x)x))/(Delta x)=`
`lim_(Delta x->0)(x-(x+Delta x))/((x+Delta x)x cdot Delta x)=lim_(Delta x->0)(-Delta x)/((x+Delta x)x cdot Delta x)=`
`=lim_(Delta x->0)-1/((x+Delta x)x)=-1/(x^2)`
So your solution is `f'(x)=-1/x^2`
This is a special case of `(x^n)'=nx^(x-1)` for `n=-1`.