What is the derivative of $cos[sin^-1 (2w)]$ ?

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Answer 1 · 2017-02-12T20:47:18

$dy/(dw)=-(4w)/(sqrt(1-4w^2))$

Let $y=cos(sin^-1(2w))$

Let $u=sin^-1(2w)$

$sinu=2w$

$(du)/(dw)cosu=2$

$(du)/(dw)=2/cosu$

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

$cos^2u=1-sin^2u=1-(2w)^2$

$cosu=sqrt(1-4w^2)$

$(du)/(dw)=2/sqrt(1-4w^2)$

$d/(dw)cos(u)=-sin(u)xx(du)/(dw)$

$=-sin(sin^-1(2w))xx2/(sqrt(1-4w^2))$

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

What is the derivative of cos[sin^-1 (2w)]cos[sin^-1 (2w)]?