Question

find the nth derivative of(x^2+4x+1)/x^3+2x^2-x-2)​

Answer 1

Answer:

What is the nth derivative of x^2÷(x+2) (2x+3)?

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x2(x+2)(2x+3)=A+Bx+2+C2x+3

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).

Step-by-step explanation:

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