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# Find the derivative of the given function from first principle : $(-x)^{-1}$

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Let $f(x) = (-x)^{-1}$. Accordingly, $f(x+h) \large\frac{-1}{x+h}$
By first principle,
$f'(x) = \lim\limits_{h \to 0} \large\frac{f(x+h)-f(x)}{h}$
$= \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{-1}{x+h} -\bigg( \large\frac{-1}{x} \bigg) \bigg] = \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{-1}{x+h} + \large\frac{1}{x} \bigg]$
$= \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{-x+(x+h)}{x(x+h)} \bigg] = \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{-x+x+h}{x(x+h)} \bigg]$
$= \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{h}{x(x+h)} \bigg] = \lim\limits_{h \to 0} \large\frac{1}{x(x+h)} = \large\frac{1}{x.x}$$ = \large\frac{1}{x^2}$
answered Apr 11, 2014