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

Dec 30, 2015

First derivative:

Use the chain rule, which states that $\frac{d}{\mathrm{dx}} \left(\cot \left(u\right)\right) = - {\csc}^{2} \left(u\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Thus,

$f ' \left(x\right) = - {\csc}^{2} \left(3 {x}^{2} - x\right) \cdot \frac{d}{\mathrm{dx}} \left(3 {x}^{2} - x\right)$

$\implies - \left(6 x - 1\right) {\csc}^{2} \left(3 {x}^{2} - x\right)$

Second derivative:

Use product rule in conjunction with the chain rule again.

When doing chain rule with the cosecant function squared, the overriding issue will be the exponent, and then the cosecant.

$f ' ' \left(x\right) = - {\csc}^{2} \left(3 {x}^{2} - x\right) \frac{d}{\mathrm{dx}} \left(6 x - 1\right) - \left(6 x - 1\right) \frac{d}{\mathrm{dx}} \left({\csc}^{2} \left(3 {x}^{2} - x\right)\right)$

Find each derivative.

$\frac{d}{\mathrm{dx}} \left(6 x - 1\right) = 6$

$\frac{d}{\mathrm{dx}} \left({\csc}^{2} \left(3 {x}^{2} - x\right)\right) = 2 \csc \left(3 {x}^{2} - x\right) \frac{d}{\mathrm{dx}} \left(\csc \left(3 {x}^{2} - x\right)\right)$

$\textcolor{w h i t e}{s s s s}$ Recall that $\frac{d}{\mathrm{dx}} \left(\csc \left(u\right)\right) = - \csc \left(u\right) \cot \left(u\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

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

$\implies - 2 \left(6 x - 1\right) {\csc}^{2} \left(3 {x}^{2} - x\right) \cot \left(3 {x}^{2} - x\right)$

Plug both the derivatives back in to find $f ' ' \left(x\right)$.

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

This can be further simplified, if you want:

$f ' ' \left(x\right) = \left(\left(72 {x}^{2} - 24 x + 2\right) \cot \left(3 {x}^{2} - x\right) - 6\right) {\csc}^{2} \left(3 {x}^{2} - x\right)$