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If $f(\mathrm{l})=1, f^{\prime}(\mathrm{l})=3$, then the derivative of $f(f(f(x)))+(f(x))^2$ at $x=1$ is
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The correct answer is:
33
Let $y=f(f(f(x)))+(f(x))^2$
Differentiating w.r.t. $x$, we get
$$
\begin{aligned}
& \frac{d y}{d x}=f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x)+2 f(x) \cdot f^{\prime}(x) \\
& \text { At } x=1 \\
& \frac{d y}{d x}=f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1)+2 f(1) \cdot f^{\prime}(1) \\
& \frac{d y}{d x}=f^{\prime}(f(1)) \cdot f^{\prime}(1) \cdot 3+2 \cdot 1 \cdot 3=f^{\prime}(1) \cdot 3 \cdot 3+6 \\
&=9 f^{\prime}(1)+6=9 \cdot 3+6=27+6=33
\end{aligned}
$$
Differentiating w.r.t. $x$, we get
$$
\begin{aligned}
& \frac{d y}{d x}=f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x)+2 f(x) \cdot f^{\prime}(x) \\
& \text { At } x=1 \\
& \frac{d y}{d x}=f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1)+2 f(1) \cdot f^{\prime}(1) \\
& \frac{d y}{d x}=f^{\prime}(f(1)) \cdot f^{\prime}(1) \cdot 3+2 \cdot 1 \cdot 3=f^{\prime}(1) \cdot 3 \cdot 3+6 \\
&=9 f^{\prime}(1)+6=9 \cdot 3+6=27+6=33
\end{aligned}
$$
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