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What is the derivative of this function #y=sin^-1(2x)#?

Answer 1

Scarlett Allison

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

We can use here the formula for derivative of #sin^(-1)x#, which is #d/(dx)sin^(-1)x=1/sqrt(1-x^2)#

As such to find derivative #(dy)/(dx)# for #y=sin^(-1)2x# using chain rule is given by

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

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

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Answer 2

Aurora Alvarado

The derivative of the function ( y = \sin^{-1}(2x) ) is ( \frac{1}{\sqrt{1 - (2x)^2}} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.