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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

$${\mathrm{f}}^{{}^{\prime }}\left(5\right)={\lim }_{\mathrm{h}\to 0}\frac{\mathrm{f}(5+\mathrm{h})-\mathrm{f}\left(5\right)}{\mathrm{h}}={\lim }_{\mathrm{h}\to 0}\frac{\frac{2}{7(5+\mathrm{h})-4}-\frac{2}{35-4}}{\mathrm{h}}$$$$={\lim }_{\mathrm{h}\to 0}\frac{\frac{2}{31+7\mathrm{h}}-\frac{2}{31}}{\mathrm{h}}={2\lim }_{\mathrm{h}\to 0}\frac{-7\mathrm{h}}{31(31+7\mathrm{h})\mathrm{h}}$$$$=2{\lim }_{\mathrm{h}\to 0}\frac{-7}{31(31+7\mathrm{h})}=\frac{-14}{{31}^2}=-\frac{14}{961}$$$$$$

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