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

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

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$1/x^2$

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

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

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