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En
 one year ago
PLEASE HELP!
find the derivative of sin(arccosx).
En
 one year ago
PLEASE HELP! find the derivative of sin(arccosx).

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En
 one year ago
Best ResponseYou've already chosen the best response.0\[=\frac{ \cos(arccosx) }{ \sqrt{1x^2}}\] i only got this far.. please the answer is x/sqrt1x^2.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0y = sin(acos(x)) asin(y) = acos(x) ... now i gotta recall atrigs :/  u = asin(y) sin(u) = y cos(u) u' = y' u' = y'/cos(u) = y'/sqrt(1sin^2(u)) u' = y'/sqrt(1y^2)  w = acos(x) cos(w) = x sin(w) w' = 1 w' = 1/sin(w) = 1/sqrt(1cos^2(w)) w' = 1/sqrt(1x^2)  u = w u' = w' therefore asin(y) = acos(x) y'/sqrt(1y^2) = 1/sqrt(1x^2) y' = sqrt(1y^2)/sqrt(1x^2)  this is what im getting without further simplifiactions, and assuming i kept track of it all

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i wonder if: sqrt(1y^2) = x

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2@amistre64 we should simplify the expression sin(arccos x)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0sounds like something smart to do :)

En
 one year ago
Best ResponseYou've already chosen the best response.0\[=\frac{ x }{ \sqrt{1x^2}}\] is the answer from my book

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2Like thisdw:1441470217632:dw

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2There might be some additional work in justtifying the sign for any angle.

En
 one year ago
Best ResponseYou've already chosen the best response.0i dont know how it became \[=x/\sqrt{1x^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y=\sin \left( \cos^{1} x \right)=\cos \left( \cos^{1} x \right)*\frac{ 1 }{ \sqrt{1x^2} }\] \[=\frac{ x }{ \sqrt{1x^2} }\]

En
 one year ago
Best ResponseYou've already chosen the best response.0@surjithayer how come cos(arccosx) became x?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cos(arccos(x)) = x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0correction \[\frac{ dy }{ dx }=\cos \left( \cos^{1} x \right)*\frac{ 1 }{ \sqrt{1x^2} }\] =?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0arccos(x) has ratio x as an input and gives an angle, say \(\theta\). well the cos(\(\theta\)) = x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ d }{ dx }\sin \left( \cos^{1} \left( x \right) \right)=\cos \left( \cos^{1} \left( x \right) \right)\cdot \frac{ d }{ dx }\cos^{1} \left( x \right) \]\[\Rightarrow \cos \left( \cos^{1} \left( x \right) \right)\cdot \frac{ 1 }{ \sqrt{1x^2} } =x \cdot \frac{ 1 }{ \sqrt{1x^2}}=\frac{ x }{ \sqrt{1x^2} }\]
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