Question

What is the derivative of this function `y=sin^-1(x/3)`?

Answer 1

`1/(sqrt(9-x^2))`

Explanation:

I am going to assume that the `sin^-1(x)` mean `arcsin(x)` instead of `csc(x)`.

Looking at the differentiation of trigonometric functions, the derivative of `arcsin(x)` is `1/sqrt(1-x^2)`.

Proof:
If `y=arcsin(x)`, then `x=sin(y)`. Differentiating both sides with respect to `x`, we get `1=cos(y)dy/dx`. Then, `dy/dx=1/cos(y)`, which is equal to `1/sqrt(1-sin^2(y))`. But we said that `y=arcsin(x)`, so `dy/dx=1/sqrt(1-x^2)`.

In order to differentiate `arcsin(x/3)`, we use the chain rule: `dy/dx=dy/(du)*(du)/dx`, with `y=arcsin(x/3)` and `u=x/3`.

Then, `d/dx(arcsin(x/3))=d/(d(x/3))(arcsin(x/3))*d/dx(x/3)=1/sqrt(1-(x/3)^2)*1/3=1/(3sqrt(1-1/9x^2))=1/(sqrt(3^2(1-1/9x^2)))=1/(sqrt(9-x^2))`.