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

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Answer 1 · 2016-12-27T07:14:25

The answer is $=(2arcsinx)/(sqrt(1-x^2))$

We need

$(sqrtx)'=1/(2sqrtx)$

$(sinx)'=cosx$

$sin^2x+cos^2x=1$

Let, $y=(arcsinx)^2$

$sqrty=arcsinx$

$sin(sqrty)=x$

$(sin(sqrty))'=x'$

$1/(2sqrty)*cos (sqrty)dy/dx=1$

$dy/dx=2sqrty/cos(sqrty)$

$sin^2(sqrty)+cos^2(sqrty)=1$

$Cos^2(sqrty)=1-x^2$

$cos(sqrty)=sqrt(1-x^2)$

Therefore,

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

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