How do you use the limit definition of the derivative to find the derivative of $f(x)=4/(x-3)$ ?

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Answer 1 · 2016-11-01T13:22:06

The limit definition is the formula $f'(x) = lim_(h -> 0) (f(x + h) - f(x))/h$ .

$f'(x) = lim_(h->0)(4/(x + h - 3) - 4/(x - 3))/h$

This involves lots of algebra--brace yourself!

$f'(x) = lim_(h->0)((4(x - 3))/((x + h -3)(x - 3)) - (4(x + h - 3))/((x - 3)(x + h - 3)))/h$

$f'(x) = lim_(h->0) ((4x - 12 - 4x - 4h + 12)/((x + h -3)(x- 3)))/h$

$f'(x) = lim_(h->0) ((-4h)/((x + h - 3)(x - 3)))/h$

$f'(x) = lim_(h->0)(-4h)/((x + h - 3)(x - 3)(h))$

$f'(x) = lim_(h->0)(-4)/((x + h - 3)(x - 3))$

We can now substitute $h =0$ into the expression.

$f'(x) = -4/((x + 0 - 3)(x- 3))$

$f'(x) = -4/((x - 3)(x - 3))$

$f'(x) = -4/(x - 3)^2$

Hopefully this helps!

How do you use the limit definition of the derivative to find the derivative of f(x)=4/(x-3)f(x)=4/(x-3)?