Mathematics - Differentiation Question with Solution | TestHub

$\begin{array}{rl}& \text{ Let }\mathrm{h}(\mathrm{x})=\mathrm{f}\left({\mathrm{g}}^{-1}(\mathrm{x})\right)\text{ and }\mathrm{g}(0)=2\\ & \begin{array}{rl}{\mathrm{h}}^{\prime }(\mathrm{x})& ={\mathrm{f}}^{\prime }\left({\mathrm{g}}^{-1}(\mathrm{x})\right)\cdot {\left({\mathrm{g}}^{-1}(\mathrm{x})\right)}^{\prime }\\ {\mathrm{h}}^{\prime }(2)& ={\mathrm{f}}^{\prime }({\mathrm{g}}^{-1}(2)\cdot {\left({\mathrm{g}}^{-1}\right)}^{\prime }(2)\\ & ={\mathrm{f}}^{\prime }(0)\cdot {\left({\mathrm{g}}^{-1}\right)}^{\prime }(2)\end{array}\end{array}$ Now $\begin{array}{rl}& f(x)={\log }_0\left(x^2+2x+4\right)\\ & f^{\prime }(x)=\frac{2x+2}{x^2+2x+4}\\ & f^{\prime }(0)=\frac{1}{2}\\ & g(x)=\frac{4}{1+e^{-2x}},g(0)=2\\ & g^{-1}(g(x))=x\end{array}$ \begin{array}{l|l}\left(\left(\mathrm{g}^{-1}\right)^{\prime}(\mathrm{g}(\mathrm{x}))\right) \mathrm{g}^{\prime}(\mathrm{x})=1 & \mathrm{~g}(\mathrm{x})=\frac{4}{1+\mathrm{e}^{-2 \mathrm{x}}} \\\left(\mathrm{g}^{-1}\right)^{\prime}(2)=\frac{1}{\mathrm{~g}^{\prime}(0)} & \mathrm{g}^{\prime}(\mathrm{x})=\frac{8 \mathrm{e}^{-2 \mathrm{x}}}{\left(1+\mathrm{e}^{-2 \mathrm{x}}\right)^2} \\=\frac{1}{2} & \mathrm{~g}^{\prime}(0)=\frac{8}{4}=2 \\So \mathrm{h}(2)=\frac{1}{4}=0.25 &\end{array}