How do you find the derivative of $arcsin(x^2/4)$ ?

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Answer 1 · 2016-11-15T19:06:58

$d/dxarcsin(x^2/4) = (x)/(2sqrt(1-x^2/16))$

$y = arcsin(x^2/4)$

$siny = x^2/4$ ..... [1]

We can now differentiate implicitly to get:

$cos(y)dy/dx = x/2$ ..... [2]

Using the fundamental trig identity $sin^2A+cos^2A-=1$ we can write:

$sin^2(y)+cos^2(y)=1$
$:. (x^2/4)^2+cos^2(y)=1$ (from [1])
$:. cos^2(y)=1-x^2/16$
$:. cos(y)=sqrt(1-x^2/16)$

Substituting into [2] we get:

$sqrt(1-x^2/16)dy/dx = x/2$

$:. dy/dx = (x)/(2sqrt(1-x^2/16))$

How do you find the derivative of arcsin(x^2/4)arcsin(x^2/4)?