# How do you differentiate y=37-sec^3(2x)?

Nov 2, 2016

$y ' = - 6 {\sec}^{3} \left(2 x\right) \tan \left(2 x\right)$

#### Explanation:

Differentiating $y$ is determined by applying the power and chain rule

Let $u \left(x\right) = {\sec}^{3} x \mathmr{and} v \left(x\right) = 2 x$
Then $f \left(x\right)$ is a composite function of $u \left(x\right) \mathmr{and} v \left(x\right)$
Thus $f \left(x\right) = u \circ v \left(x\right)$

color(red)(f'(x)'=u'(v(x))xxv'(x)

u'(x)=??
Applying the power rule differentiation
$\textcolor{b l u e}{\left({x}^{n}\right) ' = n x ' {x}^{n - 1}}$

$u ' \left(x\right) = \left({\sec}^{3} x\right) ' = 3 \times {\sec}^{3 - 1} \times \left(\sec x\right) '$
$u ' \left(x\right) = 3 {\sec}^{2} x \times \sec x \tan x$
$u ' \left(x\right) = 3 {\sec}^{3} x \tan x$

$u ' \left(v \left(x\right)\right) = 3 {\sec}^{3} \left(v \left(x\right)\right) \tan \left(v \left(x\right)\right)$

$\textcolor{red}{u ' \left(v \left(x\right)\right) = 3 {\sec}^{3} \left(2 x\right) \tan \left(2 x\right)}$

$\textcolor{red}{v ' \left(x\right) = 2}$

color(red)(f'(x)=u'(v(x))xxv'(x)

$f ' \left(x\right) = 3 {\sec}^{3} \left(2 x\right) \tan \left(2 x\right) \times 2$
$f ' \left(x\right) = 6 {\sec}^{3} \left(2 x\right) \tan \left(2 x\right)$

$y ' = \left(37\right) ' - f ' \left(x\right)$
$y ' = 0 - f ' \left(x\right)$
$y ' = - 6 {\sec}^{3} \left(2 x\right) \tan \left(2 x\right)$