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

Sagot :

Nayefx

Answer:

[tex] - \sf \displaystyle \: \frac{ \cos(2x) }{ \sin ^{2} (2x)\sqrt{ \csc(2x) } }[/tex]

Step-by-step explanation:

we are given a derivative

[tex] \displaystyle \: \frac{d}{dx} ( \sqrt{ \csc(2x) } )[/tex]

and said to figure out the first derivative

to do so

recall chain rule:

[tex] \sf\displaystyle \: \frac{d}{dx} (f(g(x)) = \frac{d}{dg} (f(g(x)) \times \frac{d}{dx} (g)[/tex]

so we get

[tex] \displaystyle \: g(x) = \csc(2x) [/tex]

rewrite the derivative using the chain rule:

[tex] \displaystyle \: \frac{d}{dg} ( \sqrt{ g } ) \times \frac{d}{dx} ( \csc(2x) )[/tex]

use square root derivative rule to simplify:

[tex] \displaystyle \: \frac{1}{ 2\sqrt{g} } \times \frac{d}{dx} ( \csc(2x) )[/tex]

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:

[tex] \sf \displaystyle \: \frac{1}{ 2\sqrt{g} } \times \frac{d}{dn}( \csc(n) ) \times \frac{d}{dx} (2x)[/tex]

use derivative formula to simplify derivatives:

[tex] \sf \displaystyle \: \frac{1}{ 2\sqrt{g} } \times - \cot(n) \csc(n) \times 2[/tex]

substitute the value of n:

[tex] \sf \displaystyle \: \frac{1}{ 2\sqrt{g} } \times - 2\cot(2x) \csc(2x) [/tex]

substitute the value of g:

[tex] \sf \displaystyle \: \frac{1}{ 2\sqrt{ \csc(2x) } } \times - 2\cot(2x) \csc(2x) [/tex]

now we need our trigonometric skills to simplify

rewrite cot and csc:

[tex] \sf \displaystyle \: \frac{1}{ 2\sqrt{ \csc(2x) } } \times - 2 \dfrac{ \cos(2x) }{ \sin(2x) } \dfrac{1}{ \sin(2x) } [/tex]

simplify multiplication:

[tex] \sf \displaystyle \: \frac{1}{ \cancel{ \: 2}\sqrt{ \csc(2x) } } \times \cancel{- 2} \dfrac{ \cos(2x) }{ \sin ^{2} (2x) } [/tex]

simplify multiplication:

[tex] - \sf \displaystyle \: \frac{ \cos(2x) }{ \sin ^{2} (2x)\sqrt{ \csc(2x) } }[/tex]

sqdancefan

9514 1404 393

Answer:

  -cot(2x)√csc(2x)

Step-by-step 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)

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