If $\mathrm{f}(1)=1, \mathrm{f}^{\prime}(1)=3$, then the derivative of $\mathrm{f}(\mathrm{f}(\mathrm{f}(x)))+(\mathrm{f}(x))^2$ at $x=1$ is

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Aicharya · Accepted Answer

Let $y=\mathrm{f}(\mathrm{f}(\mathrm{f}(x)))+(\mathrm{f}(x))^2$ $	herefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{f}^{\prime}(\mathrm{f}(\mathrm{f}(x))) \cdot \mathrm{f}^{\prime}(\mathrm{f}(x)) \cdot \mathrm{f}^{\prime}(x)+2 \mathrm{f}(x) \mathrm{f}^{\prime}(x)$ $\begin{aligned}\left.\frac{\mathrm{d} y}{\mathrm{~d} x}ight|_{x=1} & =\mathrm{f}^{\prime}(\mathrm{f}(\mathrm{f}(1))) \cdot \mathrm{f}^{\prime}(\mathrm{f}(1)) \cdot \mathrm{f}^{\prime}(1)+2 \mathrm{f}(1) \mathrm{f}^{\prime}(1) \ & =3 \cdot 3 \cdot 3+2 \cdot 1 \cdot 3 \ & =33\end{aligned}$

If $\mathrm{f}(1)=1, \mathrm{f}^{\prime}(1)=3$, then the derivative of…

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