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Question Number 165471 by leonhard77 last updated on 02/Feb/22

$$\{\begin{array}{l}h\left(3x\right)={(\frac{2-x}{x+1}-f\left(x^3\right))}^2\\ f\left(1\right)=f{}^{\prime }\left(1\right)=2\end{array}$$$$h{}^{\prime }\left(3\right)=?$$

Commented by cortano1 last updated on 04/Feb/22

$$3h^{\prime }\left(3x\right)=2(\frac{-x+2}{x+1}-f\left(x^3\right))(\frac{-3}{{(x+1)}^2}-3x^{2}f{}^{\prime }\left(x^3\right))$$$$3h{}^{\prime }\left(3\right)=2(\frac{1}{2}-f\left(1\right))(\frac{-3}{4}-3f{}^{\prime }\left(1\right))$$$$\Rightarrow h{}^{\prime }\left(3\right)=\frac{2}{3}(\frac{1}{2}-2)(-\frac{3}{4}-6)$$$$=\frac{2}{3}(-\frac{3}{2})(-\frac{27}{4})=\frac{27}{4}$$

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