Find the derivative of F(x) = xsec^-1(x^3)?

Calculus question.

1 Answer

Relevance
  • JOHN
    Lv 7
    1 month ago

    F'(x) = sec⁻¹(x³) + x[3x²{1/(x³√(x⁶ -1))} {using d/dx(sec⁻¹x) = 1/(x√(x² -1))

    = sec⁻¹(x³) + 3/√(x⁶ -1)).

    y = sec⁻¹x

    secy = x

    secytanydy/dx = 1

    dy/dx = cosycoty

    = (1/x)(1/x)1/√(1 - 1/x²)

    = 1/[x√(x² - 1)]

    • Pope
      Lv 7
      1 month agoReport

      Same mistake I made. When I edited my answer I think it was hidden from everyone but me, so I hope you don't mind my putting the last two lines in this comment.

      sec⁻¹(x³) + 3/[x³√(1 - x⁻⁶)]
      = sec⁻¹(x³) + 3sgn(x)/√(x⁶ - 1)

      The sgn(x) factor accounts for the possibility of x being negative.

Still have questions? Get answers by asking now.