# What is the second derivative of f(x)=-sec2x-cotx ?

Jul 8, 2016

$f ' ' \left(x\right) = - 4 \sec 2 x \left\{{\sec}^{2} \left(2 x\right) + {\tan}^{2} \left(2 x\right)\right\} - 2 {\csc}^{2} x \cot x .$

#### Explanation:

$f \left(x\right) = - \sec 2 x - \cot x$
$\Rightarrow f ' \left(x\right) = \left(- \sec 2 x\right) ' - \left(\cot x\right) ' = - \sec 2 x \cdot \tan 2 x \cdot \frac{d}{\mathrm{dx}} \left(2 x\right) - \left(- {\csc}^{2} x\right)$
$= - 2 \sec 2 x \cdot \tan 2 x + {\csc}^{2} x$
$\Rightarrow f ' ' \left(x\right) = \left\{f ' \left(x\right)\right\} ' = \left(- 2 \sec 2 x \tan 2 x\right) ' + \left\{{\left(\csc x\right)}^{2}\right\} '$.
$= - 2 \left\{\sec 2 x \cdot \left(\tan 2 x\right) ' + \tan 2 x \cdot \left(\sec 2 x\right) '\right\} + 2 \csc x \cdot \left(\csc x\right) '$, ......[Product Rule & Chain Rule]
$= - 2 \left[\sec 2 x \cdot \left\{{\sec}^{2} \left(2 x\right) \cdot 2\right\} + \tan 2 x \left\{\sec 2 x \cdot \tan 2 x \cdot 2\right\}\right] + 2 \csc x \left(- \csc x \cdot \cot x\right) ,$
$= - 2 \left\{2 {\sec}^{3} \left(2 x\right) + 2 \sec 2 x \cdot {\tan}^{2} \left(2 x\right)\right\} - 2 {\csc}^{2} x \cot x ,$
$= - 4 \sec 2 x \left\{{\sec}^{2} \left(2 x\right) + {\tan}^{2} \left(2 x\right)\right\} - 2 {\csc}^{2} x \cot x .$