How would I find the derivative of arccos(x) using cos(x)?

Question

And what does the "arc" part mean?

Answer 1

arccos(x) is the inverse cosine of x. it is a function defined to do exactly the opposite of what cos(x) would do. for example, since cos(0)=1, arccos(1)=0 and since cos(pi/2)=0, arccos(0)=pi/2. here, you must use say that since y=arccos(x), x=cos(y). you must now differentiate both sides of the equation with respect to x, giving:
d(x)/dx=d(cos(y))/dx
1=d(cos(y))/dy*dy/dx
1=-sin(y)*dy/dx
solving for dy/dx, you get:
dy/dx=-1/sin(y)
substituting for y, you get:
dy/dx=-1/sin(arccos(x))
now, u must do some visualization. arccos(x) returns an angle with a cosine of x, so picture a triangle with an angle theta whose cosine is x (for example, the adjacent side has a length of x and the hypotenuse has a length of 1). now, use the pythagorean theorem to find the length of the opposite leg:
x^2+b^2=1^2
b^2=1-x^2
b=sqrt(1-x^2)
you want the sine of this angle, which is:
opp/hyp=sqrt(1-x^2)/1=sqrt(1-x^2)
if u plug this back into the equation above, u get:
dy/dx=-1/sqrt(1-x^2)
therefore, the derivative of arccos(x) is -1/sqrt(1-x^2)

Answer 2

Well, I cant derive out the explanation for you using this medium, but,

d (arccos(u)) / dx = -u' / sqrt (1 - u^2)
where u is some function of x.

arccos() is sometimes also written as cos^-1 (), for example on calculators, although it means the same thing.

Answer 3

arccos (x) is just the angle whose cosine is x (it's the inverse of the cosine function.)

cos (arccos( x )) = x (by definition of arccos (x))
D(cos(arccos(x))) = D(x)= 1
= -sin(arccos(x)) D(arccos(x)) (chain rule)

So D(arccos(x)) = 1/ -sin(arccos(x)
= -(1/ (1- cos^2(arccos(x)))^1/2
since sin(y) = (1- cos^2(y))^1/2 for
any y
= -(1/(1 - x^2)^1/2)
since cos^2(arccos(x)) = x^2