Find f''(x) if f(x) = ((x^2-1)^2)/((x^2+6)^3) f''(x)=

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pramodpandey | College Teacher | (Level 3) Valedictorian

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f(x) = ((x^2-1)^2)/((x^2+6)^3)

`f(x)=((x^2-1)^2)/((x^2+6)^3)`

`f'(x)={(x^2+6)^3(2(x^2-1)2x)-(x^2-1)^2(3(x^2+6)^2(2x))}/((x^2+6)^6)`

`` `f'(x)={(x^2+3)4x(x^2-1)-(x^2-1)^2 6x}/((x^2+6)^4)`

`f'(x)={(x^2-1)(4x^2+12-6x^2+6)x}/(x^2+6)^4`

`f'(x)={x(x^2-1)(18-2x^2)}/(x^2+6)^4`                       (i)

Taking log both side

`log(f'(x))=log(x)+log(x^2-1)+log(18-2x^2)`

`-4log(x^2+6)`       (ii)

diffierntiate (ii) with respect to x ,we have

`1/(f'(x))f''(x)=1/x+(2x)/(x^2-1)-(4x)/(18-2x^2)-(8x)/(x^2+6)`

`"f''(x)=f'(x){1/x+(2x)/(x^2-1)-(4x)/(18-2x^2)-(8x)/(x^2+6)}`

`f''(x){x(x^2-1)(18-2x^2)}/(x^2+6)^4{1/x+(2x)/(x^2-1)-(4x)/(18-2x^2)-(8x)/(x^2+6)}`

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