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

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Answer 1 · 2017-02-12T20:47:18

$\frac{d y}{d w} = - \frac{4 w}{\sqrt{1 - 4 w^{2}}}$ $dy/(dw)=-(4w)/(sqrt(1-4w^2))$

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

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

$sin u = 2 w$ $sinu=2w$

$\frac{d u}{d w} cos u = 2$ $(du)/(dw)cosu=2$

$\frac{d u}{d w} = \frac{2}{cos u}$ $(du)/(dw)=2/cosu$

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

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

$cos u = \sqrt{1 - 4 w^{2}}$ $cosu=sqrt(1-4w^2)$

$\frac{d u}{d w} = \frac{2}{\sqrt{1 - 4 w^{2}}}$ $(du)/(dw)=2/sqrt(1-4w^2)$

$\frac{d}{d w} cos (u) = - sin (u) \times \frac{d u}{d w}$ $d/(dw)cos(u)=-sin(u)xx(du)/(dw)$

$= - sin ({sin}^{- 1} (2 w)) \times \frac{2}{\sqrt{1 - 4 w^{2}}}$ $=-sin(sin^-1(2w))xx2/(sqrt(1-4w^2))$

$= - \frac{4 w}{\sqrt{1 - 4 w^{2}}}$ $=-(4w)/(sqrt(1-4w^2))$

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