# What is the second derivative of f(x)=csc(3x^2-x)?

Mar 3, 2018

$f ' \left(x\right) = - \csc \left(3 {x}^{2} - x\right) \cot \left(3 {x}^{2} - x\right) \cdot \left(6 x - 1\right)$

#### Explanation:

$f \left(x\right) = \csc \left(3 {x}^{2} - x\right)$
$f ' \left(x\right) = - \csc \left(3 {x}^{2} - x\right) \cot \left(3 {x}^{2} - x\right) \cdot \frac{d}{\mathrm{dx}} \left(3 {x}^{2} - x\right)$ (cscx derivative and chain rule )

$f ' \left(x\right) = - \csc \left(3 {x}^{2} - x\right) \cot \left(3 {x}^{2} - x\right) \cdot \left(6 x - 1\right)$

derivative of $\csc \left(x\right)$:

$\csc \left(x\right) = \frac{1}{\sin} \left(x\right)$

$\frac{d}{\mathrm{dx}} \left(\csc \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(\frac{1}{\sin} \left(x\right)\right)$

$= \frac{\sin \left(x\right) \frac{d}{\mathrm{dx}} \left(1\right) - 1 \left(\frac{d}{\mathrm{dx}} \left(\sin \left(x\right)\right)\right)}{{\sin}^{2} \left(x\right)}$

$= \frac{\sin \left(x\right) \left(0\right) - \left(\cos \left(x\right)\right)}{{\sin}^{2} \left(x\right)}$

$= \frac{- \left(\cos \left(x\right)\right)}{{\sin}^{2} \left(x\right)}$

$= - \frac{1}{\sin} \left(x\right) \cdot \cos \frac{x}{\sin} \left(x\right)$

$= - \csc \left(x\right) \cot \left(x\right)$