Question 984535
I'll do the first problem to get you started.



*[Tex \LARGE f(x) = \frac{2}{x}]



*[Tex \LARGE f(x+h) = \frac{2}{x+h}]



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*[Tex \LARGE \lim_{h \to 0}\left(\frac{f(x+h)-f(x)}{h}\right) = \lim_{h \to 0}\left(\frac{\frac{2}{x+h}-\frac{2}{x}}{h}\right)]



*[Tex \LARGE \lim_{h \to 0}\left(\frac{f(x+h)-f(x)}{h}\right) = \lim_{h \to 0}\left(\frac{\frac{2x}{x(x+h)}-\frac{2}{x}}{h}\right)]



*[Tex \LARGE \lim_{h \to 0}\left(\frac{f(x+h)-f(x)}{h}\right) = \lim_{h \to 0}\left(\frac{\frac{2x}{x(x+h)}-\frac{2(x+h)}{x(x+h)}}{h}\right)]



*[Tex \LARGE \lim_{h \to 0}\left(\frac{f(x+h)-f(x)}{h}\right) = \lim_{h \to 0}\left(\frac{\frac{2x-2(x+h)}{x(x+h)}}{h}\right)]



*[Tex \LARGE \lim_{h \to 0}\left(\frac{f(x+h)-f(x)}{h}\right) = \lim_{h \to 0}\left(\frac{\frac{2x-2x-2h}{x(x+h)}}{h}\right)]



*[Tex \LARGE \lim_{h \to 0}\left(\frac{f(x+h)-f(x)}{h}\right) = \lim_{h \to 0}\left(\frac{\frac{-2h}{x(x+h)}}{h}\right)]



*[Tex \LARGE \lim_{h \to 0}\left(\frac{f(x+h)-f(x)}{h}\right) = \lim_{h \to 0}\left(\frac{-2h}{xh(x+h)}\right)]



*[Tex \LARGE \lim_{h \to 0}\left(\frac{f(x+h)-f(x)}{h}\right) = \lim_{h \to 0}\left(\frac{-2}{x(x+h)}\right)]



*[Tex \LARGE \lim_{h \to 0}\left(\frac{f(x+h)-f(x)}{h}\right) = \frac{-2}{x(x+0)}]



*[Tex \LARGE \lim_{h \to 0}\left(\frac{f(x+h)-f(x)}{h}\right) = \frac{-2}{x^2}]



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Therefore, the derivative of f(x) is {{{-2/(x^2)}}}



Now plug in x = -2 to get {{{-2/((-2)^2)=-2/(4) = -1/2}}}



The final answer is {{{-1/2}}}



I'll let you do the other problem.