# How do you find the derivative of w=1/sinz?

Mar 7, 2017

$\frac{\mathrm{dw}}{\mathrm{dz}} = - \cos \frac{z}{{\sin}^{2} \left(z\right)}$

#### Explanation:

For the general case, the derivative quotient rule tell us:
$\textcolor{w h i t e}{\text{XXX}} \frac{d \left({f}_{x}\right)}{d \left({g}_{x}\right)} = \frac{\frac{{\mathrm{df}}_{x}}{\mathrm{dx}} \cdot {g}_{x} - \frac{{\mathrm{dg}}_{x}}{\mathrm{dx}} \cdot {f}_{x}}{{g}_{x}^{2}}$

Taking $f \left(z\right) = 1$ and $g \left(z\right) = \sin \left(z\right)$
(so ${w}_{z} = \frac{{f}_{z}}{{g}_{z}}$)
and
remembering that
$\textcolor{w h i t e}{\text{XXX}} \frac{d \sin \left(z\right)}{\mathrm{dz}} = \cos \left(z\right)$
we have
$\textcolor{w h i t e}{\text{XXX}} \frac{{\mathrm{dw}}_{z}}{\mathrm{dz}} = \frac{0 \cdot \sin \left(z\right) - \cos \left(z\right) \cdot 1}{{\sin}^{2} \left(z\right)}$

$\textcolor{w h i t e}{\text{XXXXX}} = \frac{- \cos \left(z\right)}{{\sin}^{2} \left(z\right)}$

Mar 19, 2017

$\frac{\mathrm{dw}}{\mathrm{dz}} = - \csc \left(z\right) \cot \left(z\right) = \frac{- \cos z}{{\sin}^{2} z}$

#### Explanation:

The answer below is also valid, but here is a shortcut if you can remember the identity:

$w = \frac{1}{\sin} z = \csc z$

$\therefore \frac{\mathrm{dw}}{\mathrm{dz}} = \frac{d}{\mathrm{dz}} \csc z = - \csc \left(z\right) \cot \left(z\right)$

So the answer can be written as $- \csc \left(z\right) \cot \left(z\right)$ or $\frac{- \cos z}{{\sin}^{2} z}$ since they are equivalent forms.

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