Question

First derivative of
√{cosec2x).show with full step.​

Answer 1

`- \sf \displaystyle \: ( \cos(2x) )/( \sin ^(2) (2x)√( \csc(2x) ) )`

Explanation:

we are given a derivative

`\displaystyle \: (d)/(dx) ( √( \csc(2x) ) )`

and said to figure out the first derivative

to do so

recall chain rule:

`\sf\displaystyle \: (d)/(dx) (f(g(x)) = (d)/(dg) (f(g(x)) * (d)/(dx) (g)`

so we get

`\displaystyle \: g(x) = \csc(2x)`

rewrite the derivative using the chain rule:

`\displaystyle \: (d)/(dg) ( √( g ) ) * (d)/(dx) ( \csc(2x) )`

use square root derivative rule to simplify:

`\displaystyle \: (1)/( 2√(g) ) * (d)/(dx) ( \csc(2x) )`

now we need to again use chain rule composite function derivative to simplify

where we'll take a new function n so we won't mess up two g's and we'll take 2x as n

use composite function derivative to simplify:

`\sf \displaystyle \: (1)/( 2√(g) ) * (d)/(dn)( \csc(n) ) * (d)/(dx) (2x)`

use derivative formula to simplify derivatives:

`\sf \displaystyle \: (1)/( 2√(g) ) * - \cot(n) \csc(n) * 2`

substitute the value of n:

`\sf \displaystyle \: (1)/( 2√(g) ) * - 2\cot(2x) \csc(2x)`

substitute the value of g:

`\sf \displaystyle \: (1)/( 2√( \csc(2x) ) ) * - 2\cot(2x) \csc(2x)`

now we need our trigonometric skills to simplify

rewrite cot and csc:

`\sf \displaystyle \: (1)/( 2√( \csc(2x) ) ) * - 2 ( \cos(2x) )/( \sin(2x) ) (1)/( \sin(2x) )`

simplify multiplication:

`\sf \displaystyle \: \frac{1}{ \cancel{ \: 2}√( \csc(2x) ) } * \cancel{- 2} ( \cos(2x) )/( \sin ^(2) (2x) )`

simplify multiplication:

`- \sf \displaystyle \: ( \cos(2x) )/( \sin ^(2) (2x)√( \csc(2x) ) )`

Answer 2

9514 1404 393

Answer:

-cot(2x)√csc(2x)

Explanation:

Using the chain rule, ...

(d/dx)(√u) = u'/(2√u)

Here, we have u = csc(2x), so u' = -2cot(2x)csc(2x).

Then ...

(d/dx)(√csc(2x)) = (-2cot(2x)csc(2x))/(2√csc(2x)) = -cot(2x)√csc(2x)