• En

PLEASE HELP! find the derivative of sin(arccosx).

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  • En

PLEASE HELP! find the derivative of sin(arccosx).

Mathematics
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  • En
\[=\frac{ -\cos(arccosx) }{ \sqrt{1-x^2}}\] i only got this far.. please the answer is -x/sqrt1-x^2.
y = 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(1-sin^2(u)) u' = y'/sqrt(1-y^2) ----------------- w = acos(x) cos(w) = x -sin(w) w' = 1 w' = -1/sin(w) = -1/sqrt(1-cos^2(w)) w' = -1/sqrt(1-x^2) ------------------------ u = w u' = w' therefore asin(y) = acos(x) y'/sqrt(1-y^2) = -1/sqrt(1-x^2) y' = -sqrt(1-y^2)/sqrt(1-x^2) ---------------- this is what im getting without further simplifiactions, and assuming i kept track of it all
i wonder if: sqrt(1-y^2) = x

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Other answers:

@amistre64 we should simplify the expression sin(arccos x)
sounds like something smart to do :)
  • En
\[=\frac{ -x }{ \sqrt{1-x^2}}\] is the answer from my book
Like this|dw:1441470217632:dw|
There might be some additional work in justtifying the sign for any angle.
  • En
i dont know how it became \[=-x/\sqrt{1-x^2}\]
\[y=\sin \left( \cos^{-1} x \right)=\cos \left( \cos^{-1} x \right)*\frac{ -1 }{ \sqrt{1-x^2} }\] \[=-\frac{ x }{ \sqrt{1-x^2} }\]
  • En
@surjithayer how come cos(arccosx) became x?
-cos(arccos(x)) = -x
correction \[\frac{ dy }{ dx }=\cos \left( \cos^{-1} x \right)*\frac{ -1 }{ \sqrt{1-x^2} }\] =?
arccos(x) has ratio x as an input and gives an angle, say \(\theta\). well the cos(\(\theta\)) = x
  • En
thanks i get it now :)
\[\frac{ d }{ dx }\sin \left( \cos^{-1} \left( x \right) \right)=\cos \left( \cos^{-1} \left( x \right) \right)\cdot \frac{ d }{ dx }\cos^{-1} \left( x \right) \]\[\Rightarrow \cos \left( \cos^{-1} \left( x \right) \right)\cdot \frac{ -1 }{ \sqrt{1-x^2} } =x \cdot \frac{ -1 }{ \sqrt{1-x^2}}=\frac{ -x }{ \sqrt{1-x^2} }\]

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