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Question Number 147612 by ZiYangLee last updated on 22/Jul/21

$$\mathrm{Given}\mathrm{that}f\left(x\right)=\frac{{(x^3+1)}^2\sqrt{1+x^2}}{1+\sqrt{x}}.$$$$\mathrm{By}\mathrm{using}\mathrm{logarithmatic}\mathrm{differentiation},$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}f{}^{\prime }\left(1\right).$$

Answered by mathmax by abdo last updated on 22/Jul/21

$$\mathrm{another}\mathrm{way}\mathrm{we}\mathrm{have}$$$$\ln \left(\mathrm{f}\left(\mathrm{x}\right)\right)=2\ln ({\mathrm{x}}^3+1)+\frac{1}{2}\ln (1+{\mathrm{x}}^2)-\ln (1+\sqrt{\mathrm{x}})\Rightarrow$$$$\Rightarrow \frac{{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}=\frac{{6\mathrm{x}}^2}{{\mathrm{x}}^3+1}+\frac{\mathrm{x}}{1+{\mathrm{x}}^2}-\frac{1}{2\sqrt{\mathrm{x}}(1+\sqrt{\mathrm{x}})}\Rightarrow$$$$\frac{{\mathrm{f}}^{{}^{\prime }}\left(1\right)}{\mathrm{f}\left(1\right)}=\frac{6}{2}+\frac{1}{2}-\frac{1}{4}=3+\frac{1}{4}=\frac{13}{4}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(1\right)=\frac{13}{4}\mathrm{f}\left(1\right)$$$$\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(1\right)=\frac{13}{4}\times \frac{4\sqrt{2}}{2}=\frac{13\sqrt{2}}{2}$$

Answered by mathmax by abdo last updated on 22/Jul/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{({\mathrm{x}}^3+1)}^2\sqrt{1+{\mathrm{x}}^2}}{1+\sqrt{\mathrm{x}}}\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{\{{6\mathrm{x}}^2({\mathrm{x}}^3+1)\sqrt{1+{\mathrm{x}}^2}+\frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}{({\mathrm{x}}^3+1)}^2\}(1+\sqrt{\mathrm{x}})-\left(\frac{1}{2\sqrt{\mathrm{x}}}\right){({\mathrm{x}}^3+1)}^2\sqrt{1+{\mathrm{x}}^2}}{{(1+\sqrt{\mathrm{x}})}^2}$$$$\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(1\right)=\frac{(12\sqrt{2}+\frac{1}{\sqrt{2}}\times 4)\left(2\right)-\frac{1}{2}\left(4\right)\sqrt{2}}{4}$$$$=\frac{2\left(14\sqrt{2}\right)-2\sqrt{2}}{4}=\frac{26\sqrt{2}}{4}=\frac{13\sqrt{2}}{2}$$$$$$

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