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:

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

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

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

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

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

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

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