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))](https://www.tinkutara.com/question/Q185724.png)
$$\:{g}'\left({x}\right)=\frac{\frac{\mathrm{1}}{\mathrm{3}\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }}\:.{f}\left(\mathrm{1}+{x}\right)−\sqrt[{\mathrm{3}}]{{x}}\:.{f}\:'\left(\mathrm{1}+{x}\right)}{\left[{f}\left(\mathrm{1}+{x}\right)\right]^{\mathrm{2}} } \\ $$$$\:{g}\:'\left(\mathrm{1}\right)=\frac{\frac{\mathrm{1}}{\mathrm{3}}.\left(−\mathrm{4}\right)−\mathrm{1}.\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}{\mathrm{16}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{16}}.\left(−\frac{\mathrm{8}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{6}}\right)=−\frac{\mathrm{9}}{\mathrm{16}.\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{3}}{\mathrm{32}} \\ $$
Commented by mnjuly1970 last updated on 26/Jan/23
$$\:\:\:{thanks}\:{alot}\:\:{sir}\:{cortano} \\ $$