# Question #d8431

Feb 17, 2017

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

#### Explanation:

Use the chain rule:

$\frac{d}{\mathrm{dx}} {\sec}^{3} \left(\sqrt{\cos x}\right) = \frac{d}{\mathrm{du}} {u}^{3} \cdot \frac{d}{\mathrm{dy}} \sec y \cdot \frac{d}{\mathrm{dt}} \sqrt{t} \frac{d}{\mathrm{dx}} \cos x$

where:

$u = \sec \left(\sqrt{\cos x}\right)$

$y = \sqrt{\cos x}$

$t = \cos x$

so:

$\frac{d}{\mathrm{dx}} {\sec}^{3} \left(\sqrt{\cos x}\right) = 3 {u}^{2} \cdot \sec y \tan y \cdot \frac{1}{2 \sqrt{t}} \cdot \left(- \sin x\right)$

$\frac{d}{\mathrm{dx}} {\sec}^{3} \left(\sqrt{\cos x}\right) = 3 {\sec}^{2} \left(\sqrt{\cos x}\right) \cdot \sec \left(\sqrt{\cos x}\right) \tan \left(\sqrt{\cos x}\right) \cdot \frac{1}{2 \sqrt{\cos x}} \cdot \left(- \sin x\right)$

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