Question Number 134108 by Eric002 last updated on 27/Feb/21
$$\frac{{d}}{{dx}}\left(\sqrt[{\mathrm{4}}]{\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{1}}}\right) \\ $$
Answered by Ñï= last updated on 27/Feb/21
![(d/dx)((((x^3 +1)/(x^3 −1)))^(1/4) ) =(d/dx)e^((1/4)ln((x^3 +1)/(x^3 −1))) =(d/dx)e^((1/4)ln(x^3 +1)−ln(x^3 −1)) =(((x^3 +1)/(x^3 −1)))^(1/4) [(1/4)(((3x^2 )/(x^3 +1))−((3x^2 )/(x^3 −1)))]](https://www.tinkutara.com/question/Q134109.png)
$$\frac{{d}}{{dx}}\left(\sqrt[{\mathrm{4}}]{\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{1}}}\right) \\ $$$$=\frac{{d}}{{dx}}{e}^{\frac{\mathrm{1}}{\mathrm{4}}{ln}\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{1}}} \\ $$$$=\frac{{d}}{{dx}}{e}^{\frac{\mathrm{1}}{\mathrm{4}}{ln}\left({x}^{\mathrm{3}} +\mathrm{1}\right)−{ln}\left({x}^{\mathrm{3}} −\mathrm{1}\right)} \\ $$$$=\sqrt[{\mathrm{4}}]{\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{1}}}\left[\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{{x}^{\mathrm{3}} +\mathrm{1}}−\frac{\mathrm{3}{x}^{\mathrm{2}} }{{x}^{\mathrm{3}} −\mathrm{1}}\right)\right] \\ $$
Answered by EDWIN88 last updated on 28/Feb/21
$$\mathrm{let}\:\mathrm{y}\:=\:\sqrt[{\mathrm{4}}]{\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}}\: \\ $$$$\mathrm{y}^{\mathrm{4}} \:=\:\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}\:\Rightarrow\:\mathrm{ln}\:\mathrm{y}^{\mathrm{4}} \:=\:\mathrm{ln}\:\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)−\mathrm{ln}\:\left(\mathrm{x}^{\mathrm{3}} −\mathrm{1}\right) \\ $$$$\frac{\mathrm{4}}{\mathrm{y}}.\mathrm{y}'\:=\:\frac{\mathrm{3x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}\:−\:\frac{\mathrm{3x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}\:=\:\mathrm{3x}^{\mathrm{2}} \left(\frac{−\mathrm{2}}{\mathrm{x}^{\mathrm{6}} −\mathrm{1}}\right) \\ $$$$\underline{\mathrm{y}'\:=\:−\frac{\mathrm{3x}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{x}^{\mathrm{6}} −\mathrm{1}\right)}.\sqrt[{\mathrm{4}}]{\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}}\:.} \\ $$$$ \\ $$
Answered by MJS_new last updated on 28/Feb/21
![(d/dx)[h(((f(x))/(g(x))))]=h′(((f(x))/(g(x))))×(d/dx)[((f(x))/(g(x)))]= =h′(((f(x))/(g(x))))×((f′(x)g(x)−f(x)g′(x))/(g(x)^2 )) (d/dx)[(((x^3 +1)/(x^3 −1)))^(1/4) ]=(1/4)(((x^3 +1)/(x^3 −1)))^(−3/4) ×((3x^2 (x^3 −1)−(x^3 +1)3x^2 )/((x^3 −1)^2 ))= =(1/4)(((x^3 −1)/(x^3 +1)))^(3/4) ×((−6x^2 )/((x^3 −1)^2 ))= =−((3x^2 )/(2(x^3 +1)^(3/4) (x^3 −1)^(5/4) ))](https://www.tinkutara.com/question/Q134184.png)
$$\frac{{d}}{{dx}}\left[{h}\left(\frac{{f}\left({x}\right)}{{g}\left({x}\right)}\right)\right]={h}'\left(\frac{{f}\left({x}\right)}{{g}\left({x}\right)}\right)×\frac{{d}}{{dx}}\left[\frac{{f}\left({x}\right)}{{g}\left({x}\right)}\right]= \\ $$$$={h}'\left(\frac{{f}\left({x}\right)}{{g}\left({x}\right)}\right)×\frac{{f}'\left({x}\right){g}\left({x}\right)−{f}\left({x}\right){g}'\left({x}\right)}{{g}\left({x}\right)^{\mathrm{2}} } \\ $$$$\frac{{d}}{{dx}}\left[\left(\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{1}}\right)^{\mathrm{1}/\mathrm{4}} \right]=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{1}}\right)^{−\mathrm{3}/\mathrm{4}} ×\frac{\mathrm{3}{x}^{\mathrm{2}} \left({x}^{\mathrm{3}} −\mathrm{1}\right)−\left({x}^{\mathrm{3}} +\mathrm{1}\right)\mathrm{3}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{3}} −\mathrm{1}\right)^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{x}^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{3}} +\mathrm{1}}\right)^{\mathrm{3}/\mathrm{4}} ×\frac{−\mathrm{6}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{3}} −\mathrm{1}\right)^{\mathrm{2}} }= \\ $$$$=−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{4}} \left({x}^{\mathrm{3}} −\mathrm{1}\right)^{\mathrm{5}/\mathrm{4}} } \\ $$