Question

What is the derivative of `sin(arccos x)`?

Answer 1

`(dy)/(dx)=-(cos(arc cosx))/sqrt(1-x^2)=-x/sqrt(1-x^2)`

Explanation:

Here,

`y=sin(arc cosx)`

Let ,

`y=sinu and u=arc cosx`

`:.(dy)/(du)=cosu and (du)/(dx)=-1/sqrt(1-x^2)`

Using Chain Rule :

`color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dx)`

`:.(dy)/(dx)=cosu ((-1)/sqrt(1-x^2))=-cosu/sqrt(1-x^2)`

Subst. back `u=arc cosx`

`:.(dy)/(dx)=-(cos(arc cosx))/sqrt(1-x^2)=-x/sqrt(1-x^2)`

Answer 2

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

Explanation:

Here,

`y=sin(arc cosx)`

`:.y=sin(arcsin(sqrt(1-x^2))`

`:.y=sqrt(1-x^2)`

`:.(dy)/(dx)=1/(2sqrt(1-x^2))d/(dx)(1-x^2)`

`:.(dy)/(dx)=1/(2sqrt(1-x^2))(-2x)`

`:.(dy)/(dx)=-x/sqrt(1-x^2)`

Answer 3

`d/(dx) sin(arccos(x)) = -x/sqrt(1-x^2)`

Explanation:

Note that for `x in [-1, 1]` we have `arccos(x) in [0, pi]` and `sin(arccos(x)) >= 0`.

So using `cos^2 theta + sin^2 theta = 1` we have:

`sin(arccos(x)) = sqrt(1-x^2) = (1-x^2)^(1/2)`

Hence:

`d/(dx) sin(arccos(x)) = d/(dx) (1-x^2)^(1/2)`

`color(white)(d/(dx) sin(arccos(x))) = 1/2 (1-x^2)^(-1/2) * d/(dx) (1-x^2)`

`color(white)(d/(dx) sin(arccos(x))) = 1/2 (1-x^2)^(-1/2) * (-2x)`

`color(white)(d/(dx) sin(arccos(x))) = -x/sqrt(1-x^2)`