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

Calculus question.

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- JOHNLv 79 months 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)]

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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.