How do you find the derivative of arcsin sqrt(2x)arcsin√2x?

1 Answer
Jim G.
Aug 10, 2017

sqrt2/(sqrt(4x(1-2x))√2√4x(1−2x)

Explanation:

•color(white)(x)d/dx(sin^-1x)=1/sqrt(1-x^2)∙xddx(sin−1x)=1√1−x2

•color(white)(x)d/dx(sin^-1(f(x)))=1/sqrt(1-(f(x))^2)xxf'(x)

rArrd/dx(sin^-1(sqrt(2x)))

=1/sqrt(1-2x)xxd/dx(sqrt2x^(1/2))

=1/sqrt(1-2x)xx1/2sqrt2x^(-1/2)

=sqrt2/(sqrt(4x(1-2x))

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