How do you differentiate #y = arcsin(x/2)#?

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Answer 1 · 2015-06-18T13:13:52

Apply the chain rule to the derivative of #arcsin#.

You may want a more full treatment of Differentiating Inverse Sine

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

Applying the chain rule, we get:

#d/dx (arcsinu) = 1/sqrt(1-u^2) (du)/dx#

In this question #u = x/2#, so #(du)/dx = 1/2#.

We get

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

We're done with calculus, but this can be 'cleaned up' algebraically:

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

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

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

#= 1/(2 sqrt (4-x^2)/sqrt4)#

#= 1/(cancel(2) sqrt (4-x^2)/cancel(sqrt4))#

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