# How do you differentiate f(x)=sqrtcsc(2x -4)  using the chain rule?

Feb 9, 2018

$f ' \left(x\right) = - \cos \left(2 x - 4\right) \csc {\left(2 x - 4\right)}^{\frac{3}{2}}$

#### Explanation:

rewrite the function like this
$\csc {\left(2 x - 4\right)}^{\frac{1}{2}}$

use chain rule by first using power rule on the $\frac{1}{2}$ exponent,
then on the csc function, and finally on the function inside the csc

$\cancel{\frac{1}{2}} \csc {\left(2 x - 4\right)}^{- \frac{1}{2}} \cdot \left(- \csc \left(2 x - 4\right) \cot \left(2 x - 4\right)\right) \cancel{2}$

then

$\csc {\left(2 x - 4\right)}^{- \frac{1}{2}} = \frac{1}{\sqrt{\csc \left(2 x - 4\right)}}$

multiply

(-csc(2x-4)cot(2x-4))/(sqrt(csc(2x-4))

convert everything into sin and cos
$\frac{\left(- \frac{1}{\sin} \left(2 x - 4\right)\right) \cdot \left(\cos \frac{2 x - 4}{\sin} \left(2 x - 4\right)\right)}{\csc} {\left(2 x - 4\right)}^{\frac{1}{2}}$

simplify

$\frac{\frac{- \cos \left(2 x - 4\right)}{{\sin}^{2} \left(2 x - 4\right)}}{\frac{1}{\sin} {\left(2 x - 4\right)}^{\frac{1}{2}}}$

flip and multiply

$\frac{- \cos \left(2 x - 4\right) \sin {\left(2 x - 4\right)}^{\frac{1}{2}}}{\sin} ^ 2 \left(2 x - 4\right)$

subtract sin exponents

$- \cos \left(2 x - 4\right) \cdot \sin {\left(2 x - 4\right)}^{- \frac{3}{2}}$

more simplification

$- \cos \left(2 x - 4\right) \cdot \left(\frac{1}{\csc} {\left(2 x - 4\right)}^{- \frac{3}{2}}\right)$

$- \cos \left(2 x - 4\right) \csc {\left(2 x - 4\right)}^{\frac{3}{2}}$