2024-2025 MATEMÀTIQUES III (20204007) Course Materials & Test Answers | campusvirtual.urv.cat

La primera derivada de $P_n(x) = a_0 + a_1 \, (x-a) + a_2 \, (x-a)^2 + \dots + a_n\, (x-a)^n$ P_n(x) = a_0 + a_1 \, (x-a) + a_2 \, (x-a)^2 + \dots + a_n\, (x-a)^n respecte de x, és

$\displaystyle \frac{d P_n}{dx^2}= a_1$ \displaystyle \frac{d P_n}{dx^2}= a_1

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$\displaystyle \frac{d P_n}{dx^2}= a_1 + 2a_2\, (x-a) + 3a_3 \, (x-a)^2 + \dots + n a_n \, (x-a) ^{n-1}$ \displaystyle \frac{d P_n}{dx^2}= a_1 + 2a_2\, (x-a) + 3a_3 \, (x-a)^2 + \dots + n a_n \, (x-a) ^{n-1}

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$\displaystyle \frac{d P_n}{dx^2}= 2a_2 + 3\cdot 2 a_3 \, (x-a) + \dots + n (n-1) a_n \, (x-a) ^{n-2}$ \displaystyle \frac{d P_n}{dx^2}= 2a_2 + 3\cdot 2 a_3 \, (x-a) + \dots + n (n-1) a_n \, (x-a) ^{n-2}

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$\displaystyle \frac{d P_n}{dx^2}= 2a_2$ \displaystyle \frac{d P_n}{dx^2}= 2a_2

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cap de les altres és correcta

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2024-2025 MATEMÀTIQUES III (20204007)

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