How do you differentiate f(x)=sqrt(1/csc(2x -4)  using the chain rule?

Nov 29, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cot \left(2 x - 4\right) \csc \left(2 x - 4\right)}{\csc \left(2 x - 4\right)} ^ \left(\frac{3}{2}\right)$

Explanation:

First of all, this can be written as $f \left(x\right) = \frac{1}{\sqrt{\csc \left(2 x - 4\right)}}$, which in turn can be written as $f \left(x\right) = {\left(\csc \left(2 x - 4\right)\right)}^{- \frac{1}{2}}$.

To differentiate this function, we will apply the chain rule--twice.

We let $y = {u}^{- \frac{1}{2}} , u = \csc v$ and $v = \left(2 x - 4\right)$.

By the chain rule, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dv}} \times \frac{\mathrm{dv}}{\mathrm{dx}}$, therefore we will have to differentiate each function.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\frac{\mathrm{dy}}{\mathrm{du}} = - \frac{1}{2} {u}^{- \frac{3}{2}} = - \frac{1}{2 {u}^{\frac{3}{2}}}$

$\frac{\mathrm{du}}{\mathrm{dv}} = \frac{0 \times \sin v - \cos v \times 1}{\sin v} ^ 2 = - \cos \frac{v}{\sin} ^ 2 v = - \cot v \csc v$

$\frac{\mathrm{dv}}{\mathrm{dx}} = 2$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dv}} \times \frac{\mathrm{dv}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2 {u}^{\frac{3}{2}}} \times - \cot v \csc v \times 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \cot \left(2 x - 4\right) \csc \left(2 x - 4\right)}{2 {\left(\csc \left(2 x - 4\right)\right)}^{\frac{3}{2}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cot \left(2 x - 4\right) \csc \left(2 x - 4\right)}{\csc \left(2 x - 4\right)} ^ \left(\frac{3}{2}\right)$

Hopefully this helps!