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.

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

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

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

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