Find the derivative of 1/x^2 from the first principle.

Question

Find the derivative of $1/x^2$ from the first principle.

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Answer

Here ${\text{f}}(x) = \frac{1}{{{x^2}}}$
Then ${\text{f}}(x + h) = \frac{1}{{{{(x + h)}^2}}}$
We know that ${\text{f'}}(x) = \mathop {\lim }\limits_{x \to 0} \frac{{f(x + h) - f(x)}}{h}$
$\Rightarrow \;{\text{f'}}(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{{{(x + h)}^2}}} - \frac{1}{{{x^2}}}}}{h}$$= \mathop {\lim }\limits_{h \to 0} \frac{{{x^2} - {{(x + h)}^2}}}{{h{x^2}{{(x + h)}^2}}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{{x^2} - {x^2} - {h^2} - 2xh}}{{h{x^2}{{(x + h)}^2}}} = \mathop {\lim }\limits_{h \to 0} \frac{{h( - h - 2x)}}{{h{x^2}{{(x + h)}^2}}}$
$ = \frac{{ - 2x}}{{{x^2} \times {x^2}}} = \frac{{ - 2}}{{{x^3}}}$

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