How do you differentiate given y=(secx^3)sqrt(sin2x)?

May 16, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sec {x}^{3} \left(\frac{\cos 2 x}{\sqrt{\sin 2 x}} + 3 {x}^{2} \tan {x}^{3} \sqrt{\sin 2 x}\right)$

Explanation:

We have $y = u v$ where $u$ and $v$ are both functions of $x$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = u v ' + v u '$

$u = \sec {x}^{3}$
$u ' = 3 {x}^{2} \sec {x}^{3} \tan {x}^{3}$

$v = {\left(\sin 2 x\right)}^{\frac{1}{2}}$
$v ' = {\left(\sin 2 x\right)}^{- \frac{1}{2}} / 2 \cdot \frac{d}{\mathrm{dx}} \left[\sin 2 x\right] = {\left(\sin 2 x\right)}^{- \frac{1}{2}} / 2 \cdot 2 \cos 2 x = \frac{\cos 2 x}{\sqrt{\sin 2 x}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sec {x}^{3} \cos 2 x}{\sqrt{\sin 2 x}} + 3 {x}^{2} \sec {x}^{3} \tan {x}^{3} \sqrt{\sin 2 x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sec {x}^{3} \left(\frac{\cos 2 x}{\sqrt{\sin 2 x}} + 3 {x}^{2} \tan {x}^{3} \sqrt{\sin 2 x}\right)$