What is the derivative of $arcsin(x^2/4)$ ?

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Answer 1 · 2015-10-11T02:15:33

$(2x)/sqrt(16-x^4)$ for $-2 < x < 2$

Use the Chain Rule, along with the fact that $d/dx(arcsin(x))=1/sqrt(1-x^2)$ to get:

$d/dx(arcsin(x^2/4))=1/sqrt(1-(x^2/4)^2) * d/dx(x^2/4)$

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

$=(2x)/sqrt(16-x^4)$ for $-2 < x < 2$

What is the derivative of arcsin(x^2/4)arcsin(x^2/4)?