How do you use the definition of a derivative to show that if #f(x)=1/x# then #f'(x)=-1/x^2#?

Question

Ezekiel Alexander · Accepted Answer

This is a proof

By definition: # f'(x) = lim_(h->0)(f(x+h)-f(x))/h # so, with # f(x)=1/x # we have: # f'(x) = lim_(h->0)((1/(x+h)-1/x))/h# # = lim_(h->0)((x-(x-h))/(x(x+h)))/h# # = lim_(h->0)((x-x-h)/(x(x+h)))/h# # = lim_(h->0)((-h)/(x(x+h)))/h# # = lim_(h->0)(-1)/(x(x+h))# # = lim_(h->0)(-1)/(x^2+xh)# # = -1/(x^2)# QED

Samantha Adkison · Answer

undefined To use the definition of a derivative to show that if $ f(x) = \frac{1}{x} $, then $ f'(x) = -\frac{1}{x^2} $, we start with the definition of the derivative:

$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$

Substituting $ f(x) = \frac{1}{x} $ into this definition, we get:

$$ f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h} $$

Now, we need to simplify this expression. To do so, we find a common denominator:

$$ f'(x) = \lim_{h \to 0} \frac{x - (x+h)}{x(x+h)h} $$

$$ f'(x) = \lim_{h \to 0} \frac{x - x - h}{x(x+h)h} $$

$$ f'(x) = \lim_{h \to 0} \frac{-h}{x(x+h)h} $$

Now, we cancel out $ h $ from the numerator and denominator:

$$ f'(x) = \lim_{h \to 0} \frac{-1}{x(x+h)} $$

As $ h $ approaches 0, $ x+h $ approaches $ x $, so we can substitute $ x $ for $ x+h $:

$$ f'(x) = \frac{-1}{x \cdot x} $$

$$ f'(x) = -\frac{1}{x^2} $$

Thus, we have shown that if $ f(x) = \frac{1}{x} $, then $ f'(x) = -\frac{1}{x^2} $ using the definition of a derivative.