Find the derivative of f(x) = frac{1}{x}. | vedclass

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(A) We have the definition of the derivative as $f^{\prime}(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$.Substituting $f(x) = \frac{1}{x}$,we get:$f^{

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$-\frac{1}{x^{2}}$

Vedclass · Answer

(A) We have the definition of the derivative as $f^{\prime}(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$.Substituting $f(x) = \frac{1}{x}$,we get:$f^{\prime}(x) = \lim_{h 	o 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}$Taking the common denominator in the numerator:$f^{\prime}(x) = \lim_{h 	o 0} \frac{1}{h} \left[ \frac{x - (x+h)}{x(x+h)} ight]$Simplifying the expression:$f^{\prime}(x) = \lim_{h 	o 0} \frac{1}{h} \left[ \frac{-h}{x(x+h)} ight]$Canceling $h$ and evaluating the limit as $h 	o 0$:$f^{\prime}(x) = \lim_{h 	o 0} \frac{-1}{x(x+h)} = -\frac{1}{x^{2}}$.

Find the derivative of $f(x) = \frac{1}{x}$.

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