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Question Number 166829 by mnjuly1970 last updated on 01/Mar/22

$$$$$$calculate$$$$\mathrm{I}f,f\left(x\right)=\frac{(x^2+1)\sqrt[3]{(x^2+x-2)(x^4-1)(x^2+2x-3)+16}+\sqrt{x^2+3}}{(1+x+x^2)}$$$$$$$$then,f{}^{\prime }\left(1\right)=?◼m.n$$$$$$

Answered by mathsmine last updated on 01/Mar/22

$$f\left(x\right)=\frac{(x^2+1)\sqrt[3]{g\left(x\right)}+\sqrt{x^2+3}}{(1+x+x^2)}$$$$f^{\prime }\left(x\right)=\frac{(2x\sqrt[3]{g\left(x\right)}+\frac{g^{\prime }\left(x\right)}{3g^{\frac{2}{3}}\left(x\right)}(x^2+1)+\frac{x}{\sqrt{x^2+3}})}{1+x+x^2}-\frac{(1+2x)((x^2+1)\sqrt[3]{g\left(x\right)}+\sqrt{x^2+3})}{{(1+x+x^2)}^2}$$$$f^{\prime }\left(1\right)=\frac{2\sqrt[3]{16}+2\frac{g^{\prime }\left(1\right)}{3{\left(16\right)}^{\frac{2}{9}}}+\frac{1}{2}}{3}-\frac{2\sqrt[3]{16}+2}{3}$$$$g\left(x\right)=(x^2+x-2)(x^4-1)(x^2+2x-3)+16$$$${\left(fgh\right)}^{{}^{\prime }}=f^{\prime }gh+{fg}^{\prime }h+{fgh}^{\prime }$$$$1rootof{x}^2+x-2,x^4-1,x^2+2x-3\Rightarrow$$$$g^{\prime }\left(1\right)=0$$$$f^{\prime }\left(1\right)=\frac{2\sqrt[3]{16}+\frac{1}{2}}{3}-\frac{2\sqrt[3]{16}+2}{3}=-\frac{1}{2}$$$$$$

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