Let f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''( | vedclass

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(D) Given $f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3)$.Step $1$: Find the derivatives of $f(x)$.$f'(x) = 3x^2 + 2x f'(1) + f''(2)$$f''(x) = 6x + 2f'(

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$202$

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(D) Given $f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3)$.Step $1$: Find the derivatives of $f(x)$.$f'(x) = 3x^2 + 2x f'(1) + f''(2)$$f''(x) = 6x + 2f'(1)$$f'''(x) = 6$Step $2$: Evaluate the constants.For $f'''(3)$,since $f'''(x) = 6$,we have $f'''(3) = 6$.For $f''(2)$,substitute $x=2$ into $f''(x) = 6x + 2f'(1)$:$f''(2) = 6(2) + 2f'(1) = 12 + 2f'(1)$.For $f'(1)$,substitute $x=1$ into $f'(x) = 3x^2 + 2x f'(1) + f''(2)$:$f'(1) = 3(1)^2 + 2(1)f'(1) + f''(2) = 3 + 2f'(1) + f''(2)$.Step $3$: Solve the system of equations.From $f''(2) = 12 + 2f'(1)$,substitute into $f'(1) = 3 + 2f'(1) + f''(2)$:$f'(1) = 3 + 2f'(1) + 12 + 2f'(1)$$f'(1) = 15 + 4f'(1)$$-3f'(1) = 15 \implies f'(1) = -5$.Now find $f''(2)$:$f''(2) = 12 + 2(-5) = 12 - 10 = 2$.Step $4$: Calculate $f'(10)$.$f'(x) = 3x^2 + 2x f'(1) + f''(2) = 3x^2 + 2x(-5) + 2 = 3x^2 - 10x + 2$.$f'(10) = 3(10)^2 - 10(10) + 2 = 300 - 100 + 2 = 202$.

Let $f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3)$,where $x \in R$. Then $f'(10)$ is equal to:

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