Question
Find the derivative of function (-x)-1 from first principle.

Answer

Here f(x) = (-x)-1 = $- \frac{1}{x}$
Then f(x + h) = $- \frac{1}{{x + h}}$
We know that $f{\text{'}}(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$
$\Rightarrow \;f{\text{'}}(x) = \mathop {\lim }\limits_{h \to 0} \frac{{ - \frac{1}{{x + h}} - \left( { - \frac{1}{x}} \right)}}{h}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{ - x + x + h}}{{hx(x + h)}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{h}{{hx(x + h)}} = \frac{1}{{{x^2}}}$

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