If f(1)=1 and f^{prime}(1)=5,then the derivat | vedclass

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

Vedclass · Accepted Answer

$135$

Vedclass · Answer

(C) Let $y = f(f(f(x))) + (f(x))^2$. Applying the chain rule,the derivative is: $\frac{dy}{dx} = f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x) + 2f(x) \cdot f^{\prime}(x)$. At $x=1$,we have: $\left(\frac{dy}{dx}\right)_{x=1} = f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1) + 2f(1) \cdot f^{\prime}(1)$. Given $f(1)=1$ and $f^{\prime}(1)=5$,we substitute these values: $= f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot 5 + 2(1)(5)$. Since $f(1)=1$,then $f(f(1)) = f(1) = 1$. $= f^{\prime}(1) \cdot f^{\prime}(1) \cdot 5 + 10$. $= 5 \cdot 5 \cdot 5 + 10 = 125 + 10 = 135$.

If $f(1)=1$ and $f^{\prime}(1)=5$,then the derivative of $f(f(f(x)))+(f(x))^2$ at $x=1$ is

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