Answer
$F'(x)=sec^{-1}(x^3)+\dfrac{3}{x^3\sqrt{1-\frac{1}{x^6}}}=sec^{-1}(x^3)+\dfrac{3}{\sqrt{x^6-1}}$
Work Step by Step
$F(x)=xsec^{-1}(x^3)\\
F'(x)=1\times sec^{-1}(x^3)+\frac{1}{(x^3)^2\sqrt{1-\frac{1}{(x^3)^2}}}\times 3x^2\times x\\
F'(x)=sec^{-1}(x^3)+\frac{3x^3}{x^6\sqrt{1-\frac{1}{x^6}}}\\
F'(x)=sec^{-1}(x^3)+\frac{3}{x^3\sqrt{1-\frac{1}{x^6}}}\\
F'(x)=sec^{-1}(x^3)+\frac{3}{\sqrt{x^6-1}}$
