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

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Answer 1 · 2016-12-01T23:40:30

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

Let #y=arcsin(x/2)#. Rearranging this using the properties of inverse functions yields:

#sin(y)=x/2#

Differentiate both sides with respect to #x#. The left hand side will need the chain rule.

The right hand side's derivative is #1/2# since #x/2=1/2x#.

#cos(y)*dy/dx=1/2#

We can write #cos(y)# in terms of #sin(y)#, which we know equals #x/2#. From #sin^2(y)+cos^2(y)=1#, we see that #cos(y)=sqrt(1-sin^2(y))#.

#sqrt(1-sin^2(y))*dy/dx=1/2#

Since #sin(y)=x/2#, we see that #sin^2(y)=x^2/4#.

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

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

Taking the #sqrt4# out of the denominator:

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

Multiplying both sides by #2#:

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

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

Thus the derivative of #arcsin(x/2)# is #1/sqrt(4-x^2)#.