If f(1) = 1 and f'(1) = 3,then the derivative | vedclass

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

Vedclass · Accepted Answer

$33$

Vedclass · Answer

(A) Let $y = f(f(f(x))) + (f(x))^2$.Applying the chain rule,the derivative is:$\frac{dy}{dx} = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) + 2f(x) \cdot f'(x)$.At $x = 1$,we have $f(1) = 1$ and $f'(1) = 3$.Substituting these values:$\frac{dy}{dx} = f'(f(f(1))) \cdot f'(f(1)) \cdot f'(1) + 2f(1) \cdot f'(1)$.Since $f(1) = 1$,this becomes:$\frac{dy}{dx} = f'(f(1)) \cdot f'(1) \cdot f'(1) + 2(1)(3)$.$\frac{dy}{dx} = f'(1) \cdot 3 \cdot 3 + 6$.$\frac{dy}{dx} = 3 \cdot 3 \cdot 3 + 6 = 27 + 6 = 33$.

If $f(1) = 1$ and $f'(1) = 3$,then the derivative of $f(f(f(x))) + (f(x))^2$ at $x = 1$ is:

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