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

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(B) Let $y = f(f(f(x))) + (f(x))^2$. Applying the chain rule to differentiate with respect to $x$: $\frac{dy}{dx} = f^{\prime}(f(f(x))) \cdot f^{\prim

Vedclass · Accepted Answer

$33$

Vedclass · Answer

(B) Let $y = f(f(f(x))) + (f(x))^2$. Applying the chain rule to differentiate with respect to $x$: $\frac{dy}{dx} = f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x) + 2f(x) \cdot f^{\prime}(x)$. At $x = 1$: $\frac{dy}{dx} = f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1) + 2f(1) \cdot f^{\prime}(1)$. Given $f(1) = 1$ and $f^{\prime}(1) = 3$: $\frac{dy}{dx} = f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot 3 + 2(1)(3)$. Since $f(1) = 1$,then $f(f(1)) = f(1) = 1$. $\frac{dy}{dx} = f^{\prime}(1) \cdot f^{\prime}(1) \cdot 3 + 6$. Substituting $f^{\prime}(1) = 3$: $\frac{dy}{dx} = 3 \cdot 3 \cdot 3 + 6 = 27 + 6 = 33$.

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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