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What is the derivative of this function #y=1/arcsin(2x)#?

Answer 1

Aurora Alvarado

Start by differentiating #y = arcsin(2x)#.

#y = arcsin(2x) -> siny = 2x#

Through implicit differentiation, we have:

#cosy(dy/dx) = 2#

#dy/dx = 2/(cosy)#

Apply the identity #cosy = sqrt(1 - sin^2(y))#

#dy/dx = 2/sqrt(1 - sin^2y) = 2/sqrt(1 - 4x^2) ->" since siny = 2x"#

So, now we can differentiate the entire expression using the quotient rule.

#dy/dx = (0 xx arcsin(2x) - 2/sqrt(1 - 4x^2))/(arcsin(2x))^2#

#dy/dx = -2/((sqrt(1 - 4x^2))(arcsin(2x))^2)#

Hopefully this helps!

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Answer 2

Paisley Johnson

To find the derivative of the function ( y = \frac{1}{\arcsin(2x)} ), we can use the chain rule:

[ \frac{d}{dx}\left(\frac{1}{\arcsin(2x)}\right) = -\frac{1}{(1-(2x)^2)\sqrt{1-(2x)^2}} ]

So, the derivative of the function is:

[ \frac{-1}{(1-(2x)^2)\sqrt{1-(2x)^2}} ]

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Answer from HIX Tutor

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