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Question-94769




Question Number 94769 by Hamida last updated on 20/May/20
Answered by i jagooll last updated on 21/May/20
Answered by mathmax by abdo last updated on 21/May/20
f^′ (5) =lim_(h→0)    ((f(5+h)−f(5))/h) =lim_(h→0)    (((2/(7(5+h)−4))−(2/(35−4)))/h)  =lim_(h→0)     (((2/(31+7h))−(2/(31)))/h) =2lim_(h→0)    ((−7h)/(31(31+7h)h))  =2 lim_(h→0)     ((−7)/(31(31+7h))) =((−14)/(31^2 )) =−((14)/(961))
$$\mathrm{f}^{'} \left(\mathrm{5}\right)\:=\mathrm{lim}_{\mathrm{h}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{f}\left(\mathrm{5}+\mathrm{h}\right)−\mathrm{f}\left(\mathrm{5}\right)}{\mathrm{h}}\:=\mathrm{lim}_{\mathrm{h}\rightarrow\mathrm{0}} \:\:\:\frac{\frac{\mathrm{2}}{\mathrm{7}\left(\mathrm{5}+\mathrm{h}\right)−\mathrm{4}}−\frac{\mathrm{2}}{\mathrm{35}−\mathrm{4}}}{\mathrm{h}} \\ $$$$=\mathrm{lim}_{\mathrm{h}\rightarrow\mathrm{0}} \:\:\:\:\frac{\frac{\mathrm{2}}{\mathrm{31}+\mathrm{7h}}−\frac{\mathrm{2}}{\mathrm{31}}}{\mathrm{h}}\:=\mathrm{2lim}_{\mathrm{h}\rightarrow\mathrm{0}} \:\:\:\frac{−\mathrm{7h}}{\mathrm{31}\left(\mathrm{31}+\mathrm{7h}\right)\mathrm{h}} \\ $$$$=\mathrm{2}\:\mathrm{lim}_{\mathrm{h}\rightarrow\mathrm{0}} \:\:\:\:\frac{−\mathrm{7}}{\mathrm{31}\left(\mathrm{31}+\mathrm{7h}\right)}\:=\frac{−\mathrm{14}}{\mathrm{31}^{\mathrm{2}} }\:=−\frac{\mathrm{14}}{\mathrm{961}} \\ $$$$ \\ $$

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