given (f(x)=\frac{2}{x}), find (f(x)) using the limit definition of the derivative.\n\n(f(x)=)\n\nquestion…

Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative is $f^{\prime}(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=\frac{2}{x}$, then $f(x + h)=\frac{2}{x + h}$.

Step2: Substitute into the formula

$\frac{f(x + h)-f(x)}{h}=\frac{\frac{2}{x + h}-\frac{2}{x}}{h}=\frac{\frac{2x-2(x + h)}{x(x + h)}}{h}$.

Step3: Simplify the numerator

$2x-2(x + h)=2x-2x-2h=-2h$. So, $\frac{\frac{2x - 2(x + h)}{x(x + h)}}{h}=\frac{\frac{-2h}{x(x + h)}}{h}$.

Step4: Simplify the complex - fraction

$\frac{\frac{-2h}{x(x + h)}}{h}=\frac{-2h}{x(x + h)}\cdot\frac{1}{h}=-\frac{2}{x(x + h)}$.

Step5: Take the limit as $h\rightarrow0$

$f^{\prime}(x)=\lim_{h\rightarrow0}-\frac{2}{x(x + h)}$. As $h\rightarrow0$, we get $f^{\prime}(x)=-\frac{2}{x^{2}}$.

Answer:

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