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

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

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Answer 1 · 2018-03-09T16:34:46

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

Explanation:

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

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