find the derivative of f(x)=(x+2)(x^2-1/x^2+x)

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thilina-g | College Teacher | (Level 1) Educator

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`f(x) = (x+2)[(x^2-1)/(x^2+x)]`

`f'(x) = [(x^2-1)/(x^2+x)] + (x+2)[(x^2-1)/(x^2+x)]'`

I will calculate `[(x^2-1)/(x^2+x)]'` separately.

`[(x^2-1)/(x^2+x)]' = ((x^2+x)(2x) - (x^2-1)(2x+1))/(x^2+x)^2`

`[(x^2-1)/(x^2+x)]' = (2x^3+2x^2-(2x^3+x^2-2x-1))/(x^2+x)^2`

`[(x^2-1)/(x^2+x)]' = (x^2+2x+1)/(x^2+x)^2`

`[(x^2-1)/(x^2+x)]' = (x+1)^2/(x^2(x+1)^2)`

Therefore,

`[(x^2-1)/(x^2+x)]' = 1/x^2`

 

Therefore,

`f'(x) = [(x^2-1)/(x^2+x)] + (x+2)/x^2`

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

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

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

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

`f'(x) = (x^2(x+1)+2(x+1))/(x^2(x+1))`

`f'(x) = ((x^2+2)(x+1))/(x^2(x+1))`

`f'(x) = (x^2+2)/x^2`

 


Therefore, `f'(x) = (x^2+2)/x^2`

 

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