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How do you find the derivative of #arcsin(2x^2)#?

Answer 1

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

Here, #y=sin^(-1)(2x^2)#, take , #u=2x^2# #y=sin^(-1)u# #(dy)/(du)=1/sqrt(1-u^2)and(du)/(dx)=4x# #color(red)((dy)/(dx)=(dy)/(du)*(du)/(dx))=1/sqrt(1-u^2)*4x# #=>(dy)/(dx)=1/(sqrt(1-(2x^2)^2))*4x=(4x)/sqrt(1-4x^4#

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Answer 2

William Abdalla

To find the derivative of ( \arcsin(2x^2) ), you can use the chain rule of differentiation.

[ \frac{d}{dx}[\arcsin(u)] = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} ]

In this case, ( u = 2x^2 ). Differentiating ( u ) with respect to ( x ):

[ \frac{du}{dx} = 4x ]

Substitute ( u = 2x^2 ) and ( \frac{du}{dx} = 4x ) into the chain rule formula:

[ \frac{d}{dx}[\arcsin(2x^2)] = \frac{1}{\sqrt{1-(2x^2)^2}} \cdot 4x ]

[ = \frac{4x}{\sqrt{1-4x^4}} ]

So, the derivative of ( \arcsin(2x^2) ) is ( \frac{4x}{\sqrt{1-4x^4}} ).

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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.