How do you differentiate #arcsin(2x)#?

2 Answers
Jim G.
Jul 30, 2017

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

Explanation:

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

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

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

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

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

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

Narad T.
Jul 30, 2017

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

Explanation:

Let #y=arcsin(2x)#

#siny=2x#

Differentiating with respect to #x#

#dy/dxcosy=2#

#dy/dx=2/cosy#

We know that

#sin^2y+cos^2y=1#

#cos^2y=1-sin^2y=1-4x^2#

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

Therefore,

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

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