Question

How do you find the derivative of `sin(arccosx)`?

Answer 1

Use the Chain Rule, resulting in `cos(cos^-1(x))*(-1)/sqrt(1-x^2)`

Explanation:

The Chain Rule involves breaking a function up into a function of a function, sometimes expressed as `y=f(u)` and `u=g(x)`.

Step 1. Let `y=sin(u)` and `u=cos^-1(x)`.

Step 2. Chain Rule says `dy/dx=(dy)/(du)*(du)/(dx)`.

Step 3. Apply the Chain Rule to the equations in Step 1.

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

Step 4. Get rid of the `u`'s in `(dy)/(du)`. Recall that `u=cos^-1(x)`, so then

`(dy)/(du)=cos(cos^-1(x))`

Step 5. Plug your derivatives back into Chain Rule of Step 2.

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

Step 6 is often optional, but you can then try to simplify as much as you can.

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

Answer 2

I would use the fact that `sin(arccosx) = sqrt(1-x^2)`

Explanation:

`arccosx` is an angle (or, a number) between `0` and `pi` whose cosine is `x`.
The sine of an angle (or, a number) between `0` and `pi` whose cosine is `x` is `sqrt(1-x^2)`.
(Use `sin^2 theta + cos^2 theta = 1`. So `cos theta = +-sqrt(1-sin^2 theta)`
and with `theta` in `[0, pi]`, we have `cos theta > 0`.)

`sin(arccosx) = sqrt(1-x^2)`

Now i'll use `d/dx(sqrtu) = 1/(2sqrtu) (du)/dx` to get:

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

`= 1/(2sqrt(1-x^2)) * (-2x)`

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