How do you differentiate #y=sin^-1(2x)#?

1 Answer
Jim G.
Feb 11, 2017

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

Explanation:

differentiate using the #color(blue)"chain rule"#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(d/dx[sin^-1(f(x))]=1/(sqrt(1-(f(x))^2)).f'(x))color(white)(2/2)|))) #

#y=sin^-1(2x)#

#rArrdy/dx=1/(sqrt(1-(2x)^2)).d/dx(2x)#

#color(white)(rArrdy/dx)=2/(sqrt(1-4x^2))#

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