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Question Number 64287 by aliesam last updated on 16/Jul/19

Commented by mathmax by abdo last updated on 16/Jul/19

R^′ (t) =((15.0,01 e^(−0,01t) (1+1,5e^(−0,01t) )−15(1−e^(−0,01t) )(−1,5)0,01 e^(−0,01t) )/((1+1,5 e^(−0,01t) )^2 ))  =((0,15 e^(−0,01t) (1+1,5 e^(−0,01t) )+0,15.1,5 (1−e^(−0,01t) )e^(−0,01t) )/((1+1,5 e^(−0,01t) )^2 ))  =....

$${R}^{'} \left({t}\right)\:=\frac{\mathrm{15}.\mathrm{0},\mathrm{01}\:{e}^{−\mathrm{0},\mathrm{01}{t}} \left(\mathrm{1}+\mathrm{1},\mathrm{5}{e}^{−\mathrm{0},\mathrm{01}{t}} \right)−\mathrm{15}\left(\mathrm{1}−{e}^{−\mathrm{0},\mathrm{01}{t}} \right)\left(−\mathrm{1},\mathrm{5}\right)\mathrm{0},\mathrm{01}\:{e}^{−\mathrm{0},\mathrm{01}{t}} }{\left(\mathrm{1}+\mathrm{1},\mathrm{5}\:{e}^{−\mathrm{0},\mathrm{01}{t}} \right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{0},\mathrm{15}\:{e}^{−\mathrm{0},\mathrm{01}{t}} \left(\mathrm{1}+\mathrm{1},\mathrm{5}\:{e}^{−\mathrm{0},\mathrm{01}{t}} \right)+\mathrm{0},\mathrm{15}.\mathrm{1},\mathrm{5}\:\left(\mathrm{1}−{e}^{−\mathrm{0},\mathrm{01}{t}} \right){e}^{−\mathrm{0},\mathrm{01}{t}} }{\left(\mathrm{1}+\mathrm{1},\mathrm{5}\:{e}^{−\mathrm{0},\mathrm{01}{t}} \right)^{\mathrm{2}} } \\ $$$$=.... \\ $$

Commented by aliesam last updated on 16/Jul/19

thank you sir

$${thank}\:{you}\:{sir}\: \\ $$

Commented by mathmax by abdo last updated on 17/Jul/19

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

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