Question

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

Answer 1

`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`