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

If $f(1)=1$, $f^{\prime}(1)=3$, then the derivative of $\mathrm{f}(\mathrm{f}(\mathrm{f}(x)))+(\mathrm{f}(x))^2$ at $x=1$ is
  1. 12
  2. 15
  3. 9
  4. 33

Solution

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

Asked in: MHT CET 2024 (16 May Shift 1)

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