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

Mar 2, 2017

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

#### Explanation:

$\csc x = \frac{1}{\sin} x$

so

$- {\csc}^{2} x = - \frac{1}{\sin} ^ 2 x$

Using the quotient rule, we know that

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

If we say that $f \left(x\right) = - 1$ and $g \left(x\right) = {\sin}^{2} \left(x\right)$, then

$f ' \left(x\right) = 0$

$g ' \left(x\right) = 2 \sin \left(x\right) \cos \left(x\right)$

and, putting these into the quotient rule formula,

$\frac{d}{\mathrm{dx}} f \frac{x}{g} \left(x\right) = \frac{- - 1 \cdot 2 \sin \left(x\right) \cos \left(x\right)}{\sin} ^ 4 \left(x\right)$

$= \frac{2 \cos \left(x\right)}{\sin} ^ 3 \left(x\right)$

This gives the first derivative.

The second derivative is the derivative of the first derivative, so do the whole process again.

Use new $f \left(x\right) = 2 \cos \left(x\right)$ and $g \left(x\right) = {\sin}^{3} \left(x\right)$.

$f ' \left(x\right) = - 2 \sin \left(x\right)$

$g ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({\sin}^{2} \left(x\right) \cdot \sin \left(x\right)\right) =$

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

$2 \sin \left(x\right) \cos \left(x\right) \cdot \sin \left(x\right) + {\sin}^{2} \left(x\right) \cdot \cos \left(x\right) = 3 {\sin}^{2} \left(x\right) \cos \left(x\right)$

so

$g ' \left(x\right) = 3 {\sin}^{2} \left(x\right) \cos \left(x\right)$

Now, using the formula for the quotient rule,

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

$= \frac{- 2 {\sin}^{4} \left(x\right) - 6 {\sin}^{2} \left(x\right) {\cos}^{2} \left(x\right)}{\sin} ^ 6 \left(x\right)$

$= - \frac{2}{\sin} ^ 2 \left(x\right) - \frac{6 {\cos}^{2} \left(x\right)}{\sin} ^ 4 \left(x\right)$

$= - 2 {\csc}^{2} \left(x\right) - 6 {\cot}^{2} \left(x\right) {\csc}^{2} \left(x\right)$