Find the derivative of the function f(x) = fr | vedclass

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(A) Let $f(x) = \frac{1}{x^{2}}$.By the first principle of derivatives,$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$.Substituting $f(x) = \frac{1}{

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$\frac{-2}{x^{3}}$

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(A) Let $f(x) = \frac{1}{x^{2}}$.By the first principle of derivatives,$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$.Substituting $f(x) = \frac{1}{x^{2}}$,we get:$f'(x) = \lim_{h 	o 0} \frac{\frac{1}{(x+h)^{2}} - \frac{1}{x^{2}}}{h}$$f'(x) = \lim_{h 	o 0} \frac{1}{h} \left[ \frac{x^{2} - (x+h)^{2}}{x^{2}(x+h)^{2}} ight]$$f'(x) = \lim_{h 	o 0} \frac{1}{h} \left[ \frac{x^{2} - (x^{2} + 2xh + h^{2})}{x^{2}(x+h)^{2}} ight]$$f'(x) = \lim_{h 	o 0} \frac{1}{h} \left[ \frac{-2xh - h^{2}}{x^{2}(x+h)^{2}} ight]$$f'(x) = \lim_{h 	o 0} \frac{-h(2x + h)}{h \cdot x^{2}(x+h)^{2}}$$f'(x) = \lim_{h 	o 0} \frac{-(2x + h)}{x^{2}(x+h)^{2}}$Applying the limit $h 	o 0$:$f'(x) = \frac{-(2x + 0)}{x^{2}(x+0)^{2}} = \frac{-2x}{x^{2} \cdot x^{2}} = \frac{-2x}{x^{4}} = \frac{-2}{x^{3}}$.

Find the derivative of the function $f(x) = \frac{1}{x^{2}}$ using the first principle.

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