Question

find the derivative of y=arccos(sinx)

Answer 1

cos x

Answer 2

how?

Answer 3

let u =sin(x)

y= cos^-1(u)

dy/dx = dy/du * du/dx = -1 / sqrt ( 1-u^2) * cos(x)

dy/dx = cos(x) / sqrt (1-sin^2 x ) = cos(x) /cos(x) =1

Answer 4

whoops

should be -cos(x) /cos(x) = -1

Answer 5

what are you doing when you do dy/du*du/dx=-1?

Answer 6

sorry just dy/du*du/dx

Answer 7

\[ \frac{d}{dx} \cos^{-1}(x) = \frac{-1}{\sqrt{1-x^2}}\]

Answer 8

and apply chain rule

Answer 9

\[ \frac{d}{dx} \cos^{-1}(u) = \frac{-1}{\sqrt{1-x^2}} \frac{du}{dx}\]

Answer 10

thanks!