calculus

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f(x)=1/x^2
use formal definition of derivative to find slope at x=2

  • calculus -

    f'(x) = lim(h->0) [1/(x+h)^2 - 1/x^2]/h
    ignoring the limit notation for now, we have

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

    lim(h->0) is thus -2/x^3

    at x=2, that's f'(2) = -1/4

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