`h(t) = sin(arccos(t))` Find the derivative of the function

This function is a composite one, it may be expressed as `h(t) = u(v(t)),` where `v(t) = arccos(t)` and `u(y) = sin(y).` The chain rule is applicable here, `h'(t) = u'(v(t))*v'(t).`

This gives us  `h'(t) = -cos(arccos(t))*1/sqrt(1-t^2).`

Note that `cos(arccos(t)) = t` for all `t in [-1, 1],` and the final answer is

`h'(t) = -t/sqrt(1-t^2).`

We may obtain the same result if note that `sin(arccos(t)) = sqrt(1-t^2)` for all `t in [-1, 1]`  (`arccos(t)` is...

This function is a composite one, it may be expressed as `h(t) = u(v(t)),` where `v(t) = arccos(t)` and `u(y) = sin(y).` The chain rule is applicable here, `h'(t) = u'(v(t))*v'(t).`

This gives us  `h'(t) = -cos(arccos(t))*1/sqrt(1-t^2).`

Note that `cos(arccos(t)) = t` for all `t in [-1, 1],` and the final answer is

`h'(t) = -t/sqrt(1-t^2).`

We may obtain the same result if note that `sin(arccos(t)) = sqrt(1-t^2)` for all `t in [-1, 1]`  (`arccos(t)` is non-negative and therefore square root should be taken with plus sign only).