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find f'(x) if f(x)=x^2cos^-1(ln*sqrootx)

  • Calculus -

    f(x) = x^2 cos^-1(ln(sqrt(x))
    f = uv, so f' = u'v + uv'

    f' = 2x cos^-1(ln(sqrt(x)) + x^2 (cos^-1(ln(sqrt(x))'

    Now, d/dx cos^1(u) = -1/sqrt(1-u^2) u'
    u = ln(v), so u' = 1/v v'
    v = sqrt(x), so v' = 1/2sqrt(x)

    f' = 2x cos^-1(ln(sqrt(x)) + x^2 * -1/sqrt(1-ln^2(sqrt(x)) * 1/sqrt(x) * 1/2sqrt(x)

    f' = 2x cos^-1(ln(sqrt(x)) - x/sqrt(4-ln^2(x))

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