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Question Number 185723 by mnjuly1970 last updated on 26/Jan/23

Answered by cortano1 last updated on 26/Jan/23

 g′(x)=(((1/(3 (x^2 )^(1/3) )) .f(1+x)−(x)^(1/3)  .f ′(1+x))/([f(1+x)]^2 ))   g ′(1)=(((1/3).(−4)−1.((1/6)))/(16))              =(1/(16)).(−(8/6)−(1/6))=−(9/(16.6))           =−(3/(32))

$$\:{g}'\left({x}\right)=\frac{\frac{\mathrm{1}}{\mathrm{3}\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }}\:.{f}\left(\mathrm{1}+{x}\right)−\sqrt[{\mathrm{3}}]{{x}}\:.{f}\:'\left(\mathrm{1}+{x}\right)}{\left[{f}\left(\mathrm{1}+{x}\right)\right]^{\mathrm{2}} } \\ $$$$\:{g}\:'\left(\mathrm{1}\right)=\frac{\frac{\mathrm{1}}{\mathrm{3}}.\left(−\mathrm{4}\right)−\mathrm{1}.\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}{\mathrm{16}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{16}}.\left(−\frac{\mathrm{8}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{6}}\right)=−\frac{\mathrm{9}}{\mathrm{16}.\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{3}}{\mathrm{32}} \\ $$

Commented by mnjuly1970 last updated on 26/Jan/23

   thanks alot  sir cortano

$$\:\:\:{thanks}\:{alot}\:\:{sir}\:{cortano} \\ $$

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