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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Aicharya · Accepted Answer

Let $y=f(f(f(x)))+(f(x))^2$ Differentiating w.r.t. $x$, we get $\begin{aligned} & \frac{d y}{d x}=f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x)+2 f(x) \cdot f^{\prime}(x) \ & 	ext { At } x=1 \ & \frac{d y}{d x}=f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1)+2 f(1) \cdot f^{\prime}(1) \ & \frac{d y}{d x}=f^{\prime}(f(1)) \cdot f^{\prime}(1) \cdot 3+2 \cdot 1 \cdot 3=f^{\prime}(1) \cdot 3 \cdot 3+6 \ &=9 f^{\prime}(1)+6=9 \cdot 3+6=27+6=33 \end{aligned}$

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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