# How do you differentiate f(x)= cscx twice using the quotient rule?

Mar 16, 2018

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

#### Explanation:

We have: $f \left(x\right) = \csc \left(x\right) = \frac{1}{\sin \left(x\right)}$

$R i g h t a r r o w f ' \left(x\right) = \frac{\frac{d}{\mathrm{dx}} \left(1\right) \cdot \sin \left(x\right) - \frac{d}{\mathrm{dx}} \left(\sin \left(x\right)\right) \cdot 1}{{\sin}^{2} \left(x\right)}$

$R i g h t a r r o w f ' \left(x\right) = \frac{0 \cdot \sin \left(x\right) - \cos \left(x\right) \cdot 1}{{\sin}^{2} \left(x\right)}$

$R i g h t a r r o w f ' \left(x\right) = - \frac{\cos \left(x\right)}{{\sin}^{2} \left(x\right)}$

$R i g h t a r r o w f ' \left(x\right) = - \frac{\cos \left(x\right)}{\sin \left(x\right)} \cdot \frac{1}{\sin \left(x\right)}$

$R i g h t a r r o w f ' \left(x\right) = - \cot \left(x\right) \cdot \frac{1}{\sin \left(x\right)}$

$R i g h t a r r o w f ' \left(x\right) = - \frac{\cot \left(x\right)}{\sin \left(x\right)}$

Then, let's differentiate $f ' \left(x\right)$:

$R i g h t a r r o w f ' ' \left(x\right) = - \left(\frac{\frac{d}{\mathrm{dx}} \left(\cot \left(x\right)\right) \cdot \sin \left(x\right) - \frac{d}{\mathrm{dx}} \left(\sin \left(x\right)\right) \cdot \cot \left(x\right)}{{\sin}^{2} \left(x\right)}\right)$

$R i g h t a r r o w f ' ' \left(x\right) = - \left(\frac{- {\csc}^{2} \left(x\right) \cdot \sin \left(x\right) - \cos \left(x\right) \cdot \cot \left(x\right)}{{\sin}^{2} \left(x\right)}\right)$

$R i g h t a r r o w f ' ' \left(x\right) = - \left(\frac{- \left({\csc}^{2} \left(x\right) \sin \left(x\right) + \cos \left(x\right) \cot \left(x\right)\right)}{{\sin}^{2} \left(x\right)}\right)$

$R i g h t a r r o w f ' ' \left(x\right) = \frac{{\csc}^{2} \left(x\right) \sin \left(x\right) + \cos \left(x\right) \cot \left(x\right)}{{\sin}^{2} \left(x\right)}$

$R i g h t a r r o w f ' ' \left(x\right) = \frac{{\csc}^{2} \left(x\right) \sin \left(x\right)}{{\sin}^{2} \left(x\right)} + \frac{\cos \left(x\right) \cot \left(x\right)}{{\sin}^{2} \left(x\right)}$

$R i g h t a r r o w f ' ' \left(x\right) = \frac{{\csc}^{2} \left(x\right)}{\sin \left(x\right)} + \frac{\cos \left(x\right)}{\sin \left(x\right)} \cdot \frac{\cot \left(x\right)}{\sin \left(x\right)}$

$R i g h t a r r o w f ' ' \left(x\right) = {\csc}^{2} \left(x\right) \cdot \csc \left(x\right) + \cot \left(x\right) \cdot \cot \left(x\right) \cdot \csc \left(x\right)$

$R i g h t a r r o w f ' ' \left(x\right) = {\csc}^{3} \left(x\right) + {\cot}^{2} \left(x\right) \csc \left(x\right)$

One of the Pythagorean identities is ${\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right) = 1$.

If we divide through by ${\sin}^{2} \left(x\right)$, we get:

$R i g h t a r r o w \frac{{\cos}^{2} \left(x\right)}{{\sin}^{2} \left(x\right)} + \frac{{\sin}^{2} \left(x\right)}{{\sin}^{2} \left(x\right)} = \frac{1}{{\sin}^{2} \left(x\right)}$

$R i g h t a r r o w 1 + {\cot}^{2} \left(x\right) = {\csc}^{2} \left(x\right)$

$R i g h t a r r o w {\cot}^{2} \left(x\right) = {\csc}^{2} \left(x\right) - 1$

Let's apply this rearranged identity:

$R i g h t a r r o w f ' ' \left(x\right) = {\csc}^{3} \left(x\right) + \left({\csc}^{2} \left(x\right) - 1\right) \csc \left(x\right)$

$R i g h t a r r o w f ' ' \left(x\right) = {\csc}^{3} \left(x\right) + {\csc}^{3} \left(x\right) - \csc \left(x\right)$

$\therefore f ' ' \left(x\right) = 2 {\csc}^{3} \left(x\right) - \csc \left(x\right)$