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

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Answer 1 · 2016-03-01T17:03:16

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

Use chain rule to find the derivative. The derivative of arcsin x is 1/square root of 1-x^2 and then multiply by the derivative of 2x.

Answer 2 · 2016-03-01T17:09:34

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

#\frac{d}{dx}(\arcsin (2x))#

Applying chain rule as: #\frac{df(u)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}#

Let #2x=u#

#=\frac{d}{du}(\arcsin (u)). frac{d}{dx}(2x)#

#\frac{d}{du}(\arcsin(u))# = #\frac{1}{\sqrt{1-u^2}}#
{Applying the common derivative : #\frac{d}{du}(\arcsin (u))=\frac{1}{\sqrt{1-u^2}}# }

And,
#\frac{d}{dx}(2x)=2#

Substituting back #u=2x#,

#=\frac{1}{\sqrt{1-(2x)^2}}2#

Simplifying,
#=\frac{2}{\sqrt{1-4x^2}}#