If g(x) = (x^2 + 2x + 1) cdot f(x) such that | vedclass

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(D) Given $g(x) = (x^2 + 2x + 1) \cdot f(x)$.Using the product rule for differentiation: $g'(x) = \frac{d}{dx}(x^2 + 2x + 1) \cdot f(x) + (x^2 + 2x +

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$14$

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(D) Given $g(x) = (x^2 + 2x + 1) \cdot f(x)$.Using the product rule for differentiation: $g'(x) = \frac{d}{dx}(x^2 + 2x + 1) \cdot f(x) + (x^2 + 2x + 1) \cdot f'(x)$.$g'(x) = (2x + 2) \cdot f(x) + (x^2 + 2x + 1) \cdot f'(x)$.At $x = 0$: $g'(0) = (2(0) + 2) \cdot f(0) + (0^2 + 2(0) + 1) \cdot f'(0) = 2 \cdot f(0) + 1 \cdot f'(0)$.Given $\lim_{x 	o 0} \frac{f(x) - 5}{x} = 4$, since $f(0) = 5$, this limit represents the derivative $f'(0) = 4$.Substituting the values: $g'(0) = 2(5) + 1(4) = 10 + 4 = 14$.

If $g(x) = (x^2 + 2x + 1) \cdot f(x)$ such that $f(0) = 5$ and $\lim_{x \to 0} \frac{f(x) - 5}{x} = 4$, then $g'(0) = $

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