Find the nth derivative of `(x+4)/((2x+3)(x+2))
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Considering the numerator of both sides:
⟹x2=A(x+2)(2x+3)+B(2x+3)=C(x+2)
⟹x2=x2(2A)+x(7A+2B+C)+(6A+3B+2C)
Comparing like terms on both sides:
x2 term: 2A=1⟹A=12
x term: 7A+2B+C=0⟹C=−2B−7A=−2B−72
Constant term: 6A+3B+2C=0
⟹(6∗12)+3B+2∗(−2B−72)=0
⟹3+3B−4B−7=0⟹B=−4
⟹C=(−2)∗(−4)−72=92
y=x2(x+2)(2x+3)=12−4x+2+92(2x+3)
y′=(−4).(−1).(x+2)−2+92.(−1).(2x+3)−2.2
y′′=(−4).(−1).(−2).(x+2)−3+92.(−1).(−2).(2x+3)−3.22
y′′=(−4).(−1)2.2!.(x+2)−3+9.(−1)2.2!.21.(2x+3)−3
y′n=(−4).(−1)n.n!.(x+2)−(n+1)+9.(−1)n.n!.2n−1.(2x+3)−(n+1)
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