9514 1404 393
Answer:
f'(x) = (-6x² -14x -23)/(x² +5x +2)²
f''(x) = (12x³ +42x² +138x +202)/(x² +5x +2)³
Explanation:
The applicable derivative formula is ...
d(u/v) = (v·du -u·dv)/v²
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f'(x) = ((-x² -5x -2)(4x +4) -(2x² +4x -3)(-2x -5))/(-x² -5x -2)²
f'(x) = (-4x³ -24x²-28x -8 +4x³ +18x² +14x -15)/(x² +5x +2)²
f'(x) = (-6x² -14x -23)/(x² +5x +2)²
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Similarly, the second derivative is the derivative of f'(x).
f''(x) = ((x² +5x +2)²(-12x -14) -(-6x² -14x -23)(2(x² +5x +2)(2x +5)))/(x² +5x +2)⁴
f''(x) = ((x² +5x +2)(-12x -14) +2(6x² +14x +23)(2x +5))/(x² +5x +2)³
f''(x) = (12x³ +42x² +138x +202)/(x² +5x +2)³