find derivative of (arcsecx)^3
What i've done...
= 3(sec^-1(x))^2 *(1/ (x)sqrt(x^2-1))
= 3(arcsecx)^2 / (xsqrt(x^2-1))
Let's test it out:
$\begin{align*}y & = \left(\text{arcsec } x\right) ^3 \\ y^{1/3} & = \text{arcsec } x \\ \sec y^{1/3} & = x \\ \left( \cos y^{1/3} \right)^{-1} & = x \\ \left( \cos y^{1/3} \right)^{-2} \sin y^{1/3} \dfrac{1}{3}y^{-2/3} \dfrac{dy}{dx} & = 1 \\ \dfrac{dy}{dx} & = 3y^{2/3}\cot y^{1/3} \cos y^{1/3}\end{align*}$
Now, we have $\sec y^{1/3} = \dfrac{x}{1} = \dfrac{\text{hyp}}{\text{adj}}$, so $\cot y^{1/3} = \dfrac{1}{\sqrt{x^2-1}}$ and $\cos y^{1/3} = \dfrac{1}{x}$. So, we can rewrite:
$\dfrac{dy}{dx} = 3\dfrac{ \left( \text{arcsec } x \right)^2 }{ x\sqrt{x^2-1} }$
Your answer is correct.