Inverse Trig Differentiation (1 Viewer)

[DJel]

Hi,

"Find the first derivative of y = cos^-1 (sinx) in the domain -pi <= x <= pi"

I can find the first derivative fine (1 or -1) however each solution only exists in a certain domain. I.e. "-1 for -pi/2 < x < pi/2, 1 for -pi <= x < -pi/2, pi/2 < x <=pi"

How do I determine what domains each solution lies within?

Thanks.

 

[winwinwoot]

Did you put a plus minus for when u square-rooted the cos-squared or somethin?

iono if thats necessary, so if i did the question, i would just get -1 =p

 

[DJel]

yeah the 1 and -1 comes from -cos x/sqrt(cos²x)

 

[conics2008]

"Find the first derivative of y = cos-1 (sinx) in the domain -pi <= x <= pi"

Lets start the dy/dx of that shall we..

Use u= sin(x) >> du/dx = cos(x)

now dy/du= -1/ root of 1-u^2

but dy/dx = du/dx * dy/du

therefore

dy/dx = -1/root of 1-sin^2(x) * cos(x) >> simplify -cos(x)/cos^2(x) >> -1/1 = -1

now dy/dx = -1

if you want to find the domain of cos^-1(sinx)

you must find for which values do sin(x) exsist.

sin(x) domain is all real and range is -1(<=) y (<=) 1

right but cos^-1 only exsist within the range -1 and 1

you need to find for which values thats is y values in sin(x) that is less than 1 and greater than -1 .. and therefore those y values are your domain for the function cos^-1(x)

the domain of cos^-1(sinx) is all real x and range

 

[DJel]

"-1 for -pi/2 < x < pi/2, 1 for -pi <= x < -pi/2, pi/2 < x <=pi"

These are the answers given.

 

[3unitz]

cos y = sin x

from ASTC (taking negative angles in 4th and 3rd quadrants):

1st and 4th quadrants
sin (pi/2 - y) = sin x
dy/dx = -1

2nd and 3rd quadrants
sin (y + pi/2) = sin x
dy/dx = 1

edit: should probably note that the derivate at x = pi/2, and x = -pi/2 is undefined as the left hand and right hand limits of the derivative do not equal (i.e. -1 =/= 1).

 

[3unitz]

the domain of cos^-1(sinx) is all real x and range

range is between 0 and pi