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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 \[\] $\Large\frac{1}{x^2}$.

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Let $f(x) = \large\frac{1}{x^2}$. Accordingly, from the first principle.
$ f'(x) \lim\limits_{ h \to 0}\large\frac{f(x+h)-f(x)}{h}$
$ \lim\limits_{ h \to 0}\large\frac{\Large\frac{1}{(x+h)^2}-\Large\frac{1}{x^2}}{h}$
$ = \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{x^2-(x+h)^2}{x^2(x+h)^2} \bigg]$
$ = \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{x^2-x^2-h^2-2hx}{x^2(x+h)^2} \bigg]$
$ = \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{-h^2-2hx}{x^2(x+h)^2} \bigg]$
$ = \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{-h-2x}{x^2(x+h)^2} \bigg]$
$ = \large\frac{0-2x}{x^2(x+0)^2}$$ \large\frac{-2}{x^3}$
answered Apr 7, 2014 by thanvigandhi_1

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