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

2 Answers

2 votes

Answer:


- \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) ) )

User Chenelle
by
8.4k points
3 votes

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)

User Md Sajedul Islam
by
8.0k points

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