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Question Number 134108 by Eric002 last updated on 27/Feb/21

$$\frac{d}{dx}\left(\sqrt[4]{\frac{x^3+1}{x^3-1}}\right)$$

Answered by Ñï= last updated on 27/Feb/21

$$\frac{d}{dx}\left(\sqrt[4]{\frac{x^3+1}{x^3-1}}\right)$$$$=\frac{d}{dx}{e}^{\frac{1}{4}ln\frac{x^3+1}{x^3-1}}$$$$=\frac{d}{dx}{e}^{\frac{1}{4}ln(x^3+1)-ln(x^3-1)}$$$$=\sqrt[4]{\frac{x^3+1}{x^3-1}}\left[\frac{1}{4}(\frac{3x^2}{x^3+1}-\frac{3x^2}{x^3-1})\right]$$

Answered by EDWIN88 last updated on 28/Feb/21

$$\mathrm{let}\mathrm{y}=\sqrt[4]{\frac{{\mathrm{x}}^3+1}{{\mathrm{x}}^3-1}}$$$${\mathrm{y}}^4=\frac{{\mathrm{x}}^3+1}{{\mathrm{x}}^3-1}\Rightarrow \ln {\mathrm{y}}^4=\ln ({\mathrm{x}}^3+1)-\ln ({\mathrm{x}}^3-1)$$$$\frac{4}{\mathrm{y}}.{\mathrm{y}}^{\prime }=\frac{{3\mathrm{x}}^2}{{\mathrm{x}}^3+1}-\frac{{3\mathrm{x}}^2}{{\mathrm{x}}^3-1}={3\mathrm{x}}^2\left(\frac{-2}{{\mathrm{x}}^6-1}\right)$$$$\underset{―}{{\mathrm{y}}^{\prime }=-\frac{{3\mathrm{x}}^2}{2({\mathrm{x}}^6-1)}.\sqrt[4]{\frac{{\mathrm{x}}^3+1}{{\mathrm{x}}^3-1}}.}$$$$$$

Answered by MJS_new last updated on 28/Feb/21

$$\frac{d}{dx}\left[h\left(\frac{f\left(x\right)}{g\left(x\right)}\right)\right]=h^{\prime }\left(\frac{f\left(x\right)}{g\left(x\right)}\right)\times \frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=$$$$=h^{\prime }\left(\frac{f\left(x\right)}{g\left(x\right)}\right)\times \frac{f^{\prime }\left(x\right)g\left(x\right)-f\left(x\right)g^{\prime }\left(x\right)}{g{\left(x\right)}^2}$$$$\frac{d}{dx}\left[{\left(\frac{x^3+1}{x^3-1}\right)}^{1/4}\right]=\frac{1}{4}{\left(\frac{x^3+1}{x^3-1}\right)}^{-3/4}\times \frac{3x^2(x^3-1)-(x^3+1)3x^2}{{(x^3-1)}^2}=$$$$=\frac{1}{4}{\left(\frac{x^3-1}{x^3+1}\right)}^{3/4}\times \frac{-6x^2}{{(x^3-1)}^2}=$$$$=-\frac{3x^2}{2{(x^3+1)}^{3/4}{(x^3-1)}^{5/4}}$$

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