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

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Answer 1 · 2016-10-11T15:11:08

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

$y=sin^-1(x/2)$

$=>x/2=siny$

$x=2siny$

differentiate wrt $y$

$dx/dy=2cosy$

$=> dy/dx=1/(2cosy)$

$dy/dx=1/(2sqrt(1-sin^2y))$

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

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

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

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

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

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