How do you find f'(x) using the definition of a derivative for $f(x)= 1/(x-3)$ ?

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Answer 1 · 2016-03-28T22:15:30

Please see the explanation section below.

$f(x) = 1/(x-3)$

Definition of derivative: $f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h$ .

So we have

$f'(x) = lim_(hrarr0) (1/((x+h)-3)-1/(x-3))/h$ .

If we try to evaluate the limit by substitution, we get the indeterminate form $0/0$ .

We need to rewrite. Our goal is to make the denominator no longer go to $0$ . Then we can find the limit using the quotient property of limits.

It is probably not clear to a beginning student what might work, so think about what you could do. Then see if that helps.

The smart thing to do here is to write the numerator as a single fraction.

$lim_(hrarr0) (1/((x+h)-3)-1/(x-3))/h = lim_(hrarr0) ((x-3)/((x-3)(x+h-3))-(x+h-3)/((x-3)(x+h-3)))/h$

$= lim_(hrarr0)(((x-3)-(x+h-3))/((x-3)(x+h-3)))/h$

$= lim_(hrarr0)((-h)/((x-3)(x+h-3)))/h$

If we try substitution, we still get $0/0$ , but we can reduce the fraction now.

$= lim_(hrarr0)((-h)/((x-3)(x+h-3)))/(h/1)$

$= lim_(hrarr0)(-h)/((x-3)(x+h-3))*1/h$

$= lim_(hrarr0)(-1)/((x-3)(x+h-3))$

Now the numerator clearly does not approach $0$ , so the form will not be indeterminate.

$= (-1)/((x-3)(x+0-3)) = (-1)/(x-3)^2$

That is,

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

How do you find f'(x) using the definition of a derivative for f(x)= 1/(x-3)f(x)= 1/(x-3)?