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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives
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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 by thanvigandhi_1
 

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