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

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(D) Let $g(x) = f(f(f(x))) + (f(x))^2$. We need to find $g^{\prime}(1)$. Using the chain rule,the derivative of $f(f(f(x)))$ is $f^{\prime}(f(f(x))) \

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$33$

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(D) Let $g(x) = f(f(f(x))) + (f(x))^2$. We need to find $g^{\prime}(1)$. Using the chain rule,the derivative of $f(f(f(x)))$ is $f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x)$. At $x=1$,this is $f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1)$. Given $f(1)=1$ and $f^{\prime}(1)=3$,we substitute these values: $f^{\prime}(f(f(1))) = f^{\prime}(f(1)) = f^{\prime}(1) = 3$. $f^{\prime}(f(1)) = f^{\prime}(1) = 3$. $f^{\prime}(1) = 3$. So,the derivative of $f(f(f(x)))$ at $x=1$ is $3 \cdot 3 \cdot 3 = 27$. The derivative of $(f(x))^2$ is $2f(x) \cdot f^{\prime}(x)$. At $x=1$,this is $2f(1) \cdot f^{\prime}(1) = 2(1)(3) = 6$. Therefore,$g^{\prime}(1) = 27 + 6 = 33$.

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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