Math (Derivative)

Compute derivative using the definition of the derivative.
f(x)=1+1/x^2

1+ 1/(x+h)^2 - (1+ 1/(x^2))
------ divide by h and multiply by reciprocal

1/h(x+h)^2 - 1/hx^2

Next I expanded it and got this messy huge number and multiply the bottom of each other after cancelling some terms out

(2h^2x+h^3)/(h^2x^4+2h^3x^3+h^4x^2) That's as far as I got and I'm not quite sure what I did wrong or what to do next, however I checked online and the actual derivative is -2/x^3

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  1. f´(x)= lim [ f ( x + ∆h ) - f ( x ) ] / ∆h
    ∆h-> 0

    In this case :

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

    f ( x + ∆h ) = 1 + 1 / ( x + ∆h ) ^ 2

    f ( x + ∆h ) - f ( x ) =

    1 + 1 / ( x + ∆h ) ^ 2 - ( 1 + 1 / x ^ 2 ) =

    1 + 1 / ( x + ∆h ) ^ 2 - 1 - 1 / x ^ 2 =

    1 / ( x + ∆h ) ^ 2 - 1 / x ^ 2 =

    x ^ 2 / [ x ^ 2 * ( x + ∆h ) ^ 2 ] - ( x + ∆h ) ^ 2 / [ x ^ 2 * ( x + ∆h ) ^ 2 ] =

    [ x ^ 2 - ( x + ∆h ) ^ 2 ] / [ x ^ 2 * ( x + ∆h ) ^ 2 ] =

    [ x ^ 2 - ( x ^ 2 + 2 x ∆h + ∆h ^ 2 ) ] / [ x ^ 2 * ( x + ∆h ) ^ 2 ] =

    [ x ^ 2 - x ^ 2 - 2 x ∆h - ∆h ^ 2 ] / [ x ^ 2 * ( x + ∆h ) ^ 2 ] =

    [ - 2 x ∆h - ∆h ^ 2 ] / [ x ^ 2 * ( x + ∆h ) ^ 2 ] =

    ∆h [ - 2 x - ∆h ] / [ x ^ 2 * ( x + ∆h ) ^ 2 ]

    f ( x + ∆h ) - f ( x ) = ∆h [ - 2 x - ∆h ] / [ x ^ 2 * ( x + ∆h ) ^ 2 ]

    Now:

    f´(x)= lim [ f ( x + ∆h ) - f ( x ) ] / ∆h
    ∆h-> 0

    f´ (x)= lim { ∆h [ - 2 x - ∆h ] / [ x ^ 2 * ( x + ∆h ) ^ 2 ] } / ∆h
    ∆h-> 0

    f ´ (x)= lim ∆h [ - 2 x - ∆h ] / { ∆h [ x ^ 2 * ( x + ∆h ) ^ 2 ] }
    ∆h-> 0

    f ´ (x)= lim [ - 2 x - ∆h ] / [ x ^ 2 * ( x + ∆h ) ^ 2 ]
    ∆h-> 0

    As ∆h-> 0 then:

    - 2 x - ∆h = - 2 x - 0 = - 2 x

    ( x + ∆h ) ^ 2 = ( x + 0 ) ^ 2 = x ^ 2

    so :

    f´(x)= lim [ - 2 x - ∆h ] / [ x ^ 2 * ( x + ∆h ) ^ 2 ] =
    ∆h-> 0

    - 2 x / ( x ^ 2 * x ^ 2 ) =

    - 2 x / x ^ 4 =

    - 2 x / ( x * x ^ 3 ) =

    - 2 / x ^ 3

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  2. In your homework you can replace ∆h with h

    Is all the same.

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