Question

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

Answer 1

\[=\frac{ -\cos(arccosx) }{ \sqrt{1-x^2}}\] i only got this far.. please the answer is -x/sqrt1-x^2.

Answer 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

Answer 3

i wonder if: sqrt(1-y^2) = x

Answer 4

@amistre64 we should simplify the expression sin(arccos x)

Answer 5

sounds like something smart to do :)

Answer 6

\[=\frac{ -x }{ \sqrt{1-x^2}}\] is the answer from my book

Answer 7

Like this