If f(1)=1 and f^{prime}(1)=3,then the derivat | vedclass

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(D) Let $y = f(f(f(x))) + (f(x))^2$. Applying the chain rule,we get: $\frac{dy}{dx} = f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x) +

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$33$

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(D) Let $y = f(f(f(x))) + (f(x))^2$. Applying the chain rule,we get: $\frac{dy}{dx} = f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x) + 2f(x)f^{\prime}(x)$. Evaluating at $x=1$: $\left. \frac{dy}{dx} \right|_{x=1} = f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1) + 2f(1)f^{\prime}(1)$. Given $f(1)=1$ and $f^{\prime}(1)=3$: $\left. \frac{dy}{dx} \right|_{x=1} = f^{\prime}(f(1)) \cdot f^{\prime}(1) \cdot 3 + 2(1)(3)$. Since $f(1)=1$,this becomes: $\left. \frac{dy}{dx} \right|_{x=1} = f^{\prime}(1) \cdot 3 \cdot 3 + 6 = 3 \cdot 3 \cdot 3 + 6 = 27 + 6 = 33$.

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

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