# How do you differentiate f(x)=tan((1/cos(7x))^2) using the chain rule?

Nov 10, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = 14 {\sec}^{2} \left({\sec}^{2} \left(7 x\right)\right) \tan \left(7 x\right) {\sec}^{2} \left(7 x\right)$

#### Explanation:

Okay, so, we have

$y = \tan \left({\sec}^{2} \left(7 x\right)\right)$

Call ${\sec}^{2} \left(7 x\right) = u$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{du}} \tan \left(u\right) \frac{\mathrm{du}}{\mathrm{dx}} = {\sec}^{2} \left(u\right) \frac{d}{\mathrm{dx}} {\sec}^{2} \left(7 x\right)$

Call $\sec \left(7 x\right) = v$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} \left(u\right) \frac{d}{\mathrm{dv}} {v}^{2} \frac{\mathrm{dv}}{\mathrm{dx}} = 2 v \cdot {\sec}^{2} \left(u\right) \frac{d}{\mathrm{dx}} \left(\sec \left(7 x\right)\right)$

Call $7 x = w$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 v \cdot {\sec}^{2} \left(u\right) \frac{d}{\mathrm{dw}} \sec \left(w\right) \frac{\mathrm{dw}}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 v \cdot {\sec}^{2} \left(u\right) \cdot \tan \left(w\right) \sec \left(w\right) \frac{d}{\mathrm{dx}} \left(7 x\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 14 v \cdot {\sec}^{2} \left(u\right) \cdot \tan \left(w\right) \sec \left(w\right)$

Substituting $w$ back to $x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 14 v \cdot {\sec}^{2} \left(u\right) \cdot \tan \left(7 x\right) \sec \left(7 x\right)$

Substituting $v$ back to $x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 14 \cdot {\sec}^{2} \left(u\right) \cdot \tan \left(7 x\right) {\sec}^{2} \left(7 x\right)$

Substituing $u$ back to $x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 14 {\sec}^{2} \left({\sec}^{2} \left(7 x\right)\right) \tan \left(7 x\right) {\sec}^{2} \left(7 x\right)$